YES

We show the termination of the TRS R:

  active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
  active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
  active(from(X)) -> mark(cons(X,from(s(X))))
  active(|2nd|(X)) -> |2nd|(active(X))
  active(cons(X1,X2)) -> cons(active(X1),X2)
  active(from(X)) -> from(active(X))
  active(s(X)) -> s(active(X))
  active(cons1(X1,X2)) -> cons1(active(X1),X2)
  active(cons1(X1,X2)) -> cons1(X1,active(X2))
  |2nd|(mark(X)) -> mark(|2nd|(X))
  cons(mark(X1),X2) -> mark(cons(X1,X2))
  from(mark(X)) -> mark(from(X))
  s(mark(X)) -> mark(s(X))
  cons1(mark(X1),X2) -> mark(cons1(X1,X2))
  cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
  proper(|2nd|(X)) -> |2nd|(proper(X))
  proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
  proper(from(X)) -> from(proper(X))
  proper(s(X)) -> s(proper(X))
  proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
  |2nd|(ok(X)) -> ok(|2nd|(X))
  cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
  from(ok(X)) -> ok(from(X))
  s(ok(X)) -> ok(s(X))
  cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
  top(mark(X)) -> top(proper(X))
  top(ok(X)) -> top(active(X))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(|2nd|(cons(X,X1))) -> |2nd|#(cons1(X,X1))
p2: active#(|2nd|(cons(X,X1))) -> cons1#(X,X1)
p3: active#(from(X)) -> cons#(X,from(s(X)))
p4: active#(from(X)) -> from#(s(X))
p5: active#(from(X)) -> s#(X)
p6: active#(|2nd|(X)) -> |2nd|#(active(X))
p7: active#(|2nd|(X)) -> active#(X)
p8: active#(cons(X1,X2)) -> cons#(active(X1),X2)
p9: active#(cons(X1,X2)) -> active#(X1)
p10: active#(from(X)) -> from#(active(X))
p11: active#(from(X)) -> active#(X)
p12: active#(s(X)) -> s#(active(X))
p13: active#(s(X)) -> active#(X)
p14: active#(cons1(X1,X2)) -> cons1#(active(X1),X2)
p15: active#(cons1(X1,X2)) -> active#(X1)
p16: active#(cons1(X1,X2)) -> cons1#(X1,active(X2))
p17: active#(cons1(X1,X2)) -> active#(X2)
p18: |2nd|#(mark(X)) -> |2nd|#(X)
p19: cons#(mark(X1),X2) -> cons#(X1,X2)
p20: from#(mark(X)) -> from#(X)
p21: s#(mark(X)) -> s#(X)
p22: cons1#(mark(X1),X2) -> cons1#(X1,X2)
p23: cons1#(X1,mark(X2)) -> cons1#(X1,X2)
p24: proper#(|2nd|(X)) -> |2nd|#(proper(X))
p25: proper#(|2nd|(X)) -> proper#(X)
p26: proper#(cons(X1,X2)) -> cons#(proper(X1),proper(X2))
p27: proper#(cons(X1,X2)) -> proper#(X1)
p28: proper#(cons(X1,X2)) -> proper#(X2)
p29: proper#(from(X)) -> from#(proper(X))
p30: proper#(from(X)) -> proper#(X)
p31: proper#(s(X)) -> s#(proper(X))
p32: proper#(s(X)) -> proper#(X)
p33: proper#(cons1(X1,X2)) -> cons1#(proper(X1),proper(X2))
p34: proper#(cons1(X1,X2)) -> proper#(X1)
p35: proper#(cons1(X1,X2)) -> proper#(X2)
p36: |2nd|#(ok(X)) -> |2nd|#(X)
p37: cons#(ok(X1),ok(X2)) -> cons#(X1,X2)
p38: from#(ok(X)) -> from#(X)
p39: s#(ok(X)) -> s#(X)
p40: cons1#(ok(X1),ok(X2)) -> cons1#(X1,X2)
p41: top#(mark(X)) -> top#(proper(X))
p42: top#(mark(X)) -> proper#(X)
p43: top#(ok(X)) -> top#(active(X))
p44: top#(ok(X)) -> active#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p41, p43}
  {p7, p9, p11, p13, p15, p17}
  {p25, p27, p28, p30, p32, p34, p35}
  {p18, p36}
  {p22, p23, p40}
  {p19, p37}
  {p20, p38}
  {p21, p39}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: top#(ok(X)) -> top#(active(X))
p2: top#(mark(X)) -> top#(proper(X))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        top#_A(x1) = max{0, x1 - 1}
        ok_A(x1) = x1 + 30
        active_A(x1) = x1 + 11
        mark_A(x1) = max{18, x1}
        proper_A(x1) = x1
        |2nd|_A(x1) = x1 + 28
        cons_A(x1,x2) = max{x1 + 31, x2 + 5}
        from_A(x1) = x1 + 20
        s_A(x1) = x1 + 6
        cons1_A(x1,x2) = max{x1 + 42, x2 + 12}
  
    precedence:
    
      top# = ok = active = mark = proper = |2nd| = cons = from = s = cons1
    
    partial status:
  
      pi(top#) = []
      pi(ok) = []
      pi(active) = [1]
      pi(mark) = []
      pi(proper) = [1]
      pi(|2nd|) = [1]
      pi(cons) = []
      pi(from) = [1]
      pi(s) = []
      pi(cons1) = [1, 2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: top#(mark(X)) -> top#(proper(X))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: top#(mark(X)) -> top#(proper(X))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        top#_A(x1) = max{6, x1 + 2}
        mark_A(x1) = max{7, x1 - 3}
        proper_A(x1) = max{5, x1 - 5}
        |2nd|_A(x1) = max{5, x1}
        cons_A(x1,x2) = max{2, x1}
        from_A(x1) = max{4, x1}
        s_A(x1) = max{4, x1}
        cons1_A(x1,x2) = max{x1, x2}
        ok_A(x1) = 1
  
    precedence:
    
      proper = cons1 = ok > cons = from > mark = |2nd| = s > top#
    
    partial status:
  
      pi(top#) = [1]
      pi(mark) = []
      pi(proper) = []
      pi(|2nd|) = []
      pi(cons) = []
      pi(from) = []
      pi(s) = []
      pi(cons1) = []
      pi(ok) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(cons1(X1,X2)) -> active#(X2)
p2: active#(cons1(X1,X2)) -> active#(X1)
p3: active#(s(X)) -> active#(X)
p4: active#(from(X)) -> active#(X)
p5: active#(cons(X1,X2)) -> active#(X1)
p6: active#(|2nd|(X)) -> active#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = max{4, x1 + 3}
        cons1_A(x1,x2) = max{x1 + 2, x2 + 2}
        s_A(x1) = x1
        from_A(x1) = x1
        cons_A(x1,x2) = max{x1, x2 + 1}
        |2nd|_A(x1) = x1
  
    precedence:
    
      active# = cons1 = s = from = cons = |2nd|
    
    partial status:
  
      pi(active#) = []
      pi(cons1) = [2]
      pi(s) = [1]
      pi(from) = [1]
      pi(cons) = [2]
      pi(|2nd|) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(cons1(X1,X2)) -> active#(X1)
p2: active#(s(X)) -> active#(X)
p3: active#(from(X)) -> active#(X)
p4: active#(cons(X1,X2)) -> active#(X1)
p5: active#(|2nd|(X)) -> active#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(cons1(X1,X2)) -> active#(X1)
p2: active#(|2nd|(X)) -> active#(X)
p3: active#(cons(X1,X2)) -> active#(X1)
p4: active#(from(X)) -> active#(X)
p5: active#(s(X)) -> active#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = x1 + 2
        cons1_A(x1,x2) = max{x1, x2}
        |2nd|_A(x1) = x1
        cons_A(x1,x2) = max{x1 + 2, x2 + 2}
        from_A(x1) = x1
        s_A(x1) = x1
  
    precedence:
    
      active# = cons1 = |2nd| = cons = from = s
    
    partial status:
  
      pi(active#) = [1]
      pi(cons1) = [1, 2]
      pi(|2nd|) = [1]
      pi(cons) = [1, 2]
      pi(from) = [1]
      pi(s) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(|2nd|(X)) -> active#(X)
p2: active#(cons(X1,X2)) -> active#(X1)
p3: active#(from(X)) -> active#(X)
p4: active#(s(X)) -> active#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(|2nd|(X)) -> active#(X)
p2: active#(s(X)) -> active#(X)
p3: active#(from(X)) -> active#(X)
p4: active#(cons(X1,X2)) -> active#(X1)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = max{4, x1 + 2}
        |2nd|_A(x1) = max{3, x1 + 1}
        s_A(x1) = x1
        from_A(x1) = x1 + 2
        cons_A(x1,x2) = max{x1, x2 + 1}
  
    precedence:
    
      active# = |2nd| = s = from = cons
    
    partial status:
  
      pi(active#) = []
      pi(|2nd|) = [1]
      pi(s) = [1]
      pi(from) = [1]
      pi(cons) = [2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(s(X)) -> active#(X)
p2: active#(from(X)) -> active#(X)
p3: active#(cons(X1,X2)) -> active#(X1)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(s(X)) -> active#(X)
p2: active#(cons(X1,X2)) -> active#(X1)
p3: active#(from(X)) -> active#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = x1 + 1
        s_A(x1) = x1
        cons_A(x1,x2) = max{x1, x2}
        from_A(x1) = x1
  
    precedence:
    
      active# = s = cons = from
    
    partial status:
  
      pi(active#) = [1]
      pi(s) = [1]
      pi(cons) = [1, 2]
      pi(from) = [1]

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(s(X)) -> active#(X)
p2: active#(from(X)) -> active#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(s(X)) -> active#(X)
p2: active#(from(X)) -> active#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = max{3, x1 + 2}
        s_A(x1) = x1 + 3
        from_A(x1) = max{1, x1}
  
    precedence:
    
      active# = s = from
    
    partial status:
  
      pi(active#) = [1]
      pi(s) = [1]
      pi(from) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(from(X)) -> active#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(from(X)) -> active#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = x1 + 2
        from_A(x1) = x1 + 2
  
    precedence:
    
      active# = from
    
    partial status:
  
      pi(active#) = [1]
      pi(from) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: proper#(cons1(X1,X2)) -> proper#(X2)
p2: proper#(cons1(X1,X2)) -> proper#(X1)
p3: proper#(s(X)) -> proper#(X)
p4: proper#(from(X)) -> proper#(X)
p5: proper#(cons(X1,X2)) -> proper#(X2)
p6: proper#(cons(X1,X2)) -> proper#(X1)
p7: proper#(|2nd|(X)) -> proper#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        proper#_A(x1) = max{2, x1 + 1}
        cons1_A(x1,x2) = max{x1, x2}
        s_A(x1) = x1
        from_A(x1) = x1
        cons_A(x1,x2) = max{x1, x2}
        |2nd|_A(x1) = x1 + 1
  
    precedence:
    
      proper# = cons1 = s = from = cons = |2nd|
    
    partial status:
  
      pi(proper#) = [1]
      pi(cons1) = [1, 2]
      pi(s) = [1]
      pi(from) = [1]
      pi(cons) = [1, 2]
      pi(|2nd|) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: proper#(cons1(X1,X2)) -> proper#(X1)
p2: proper#(s(X)) -> proper#(X)
p3: proper#(from(X)) -> proper#(X)
p4: proper#(cons(X1,X2)) -> proper#(X2)
p5: proper#(cons(X1,X2)) -> proper#(X1)
p6: proper#(|2nd|(X)) -> proper#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: proper#(cons1(X1,X2)) -> proper#(X1)
p2: proper#(|2nd|(X)) -> proper#(X)
p3: proper#(cons(X1,X2)) -> proper#(X1)
p4: proper#(cons(X1,X2)) -> proper#(X2)
p5: proper#(from(X)) -> proper#(X)
p6: proper#(s(X)) -> proper#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        proper#_A(x1) = max{5, x1 + 2}
        cons1_A(x1,x2) = max{x1 + 1, x2}
        |2nd|_A(x1) = max{4, x1 + 1}
        cons_A(x1,x2) = max{x1 + 1, x2}
        from_A(x1) = x1 + 3
        s_A(x1) = max{3, x1}
  
    precedence:
    
      proper# = cons1 = |2nd| = cons = from = s
    
    partial status:
  
      pi(proper#) = []
      pi(cons1) = [2]
      pi(|2nd|) = [1]
      pi(cons) = [2]
      pi(from) = [1]
      pi(s) = [1]

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: proper#(cons1(X1,X2)) -> proper#(X1)
p2: proper#(cons(X1,X2)) -> proper#(X1)
p3: proper#(cons(X1,X2)) -> proper#(X2)
p4: proper#(from(X)) -> proper#(X)
p5: proper#(s(X)) -> proper#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: proper#(cons1(X1,X2)) -> proper#(X1)
p2: proper#(s(X)) -> proper#(X)
p3: proper#(from(X)) -> proper#(X)
p4: proper#(cons(X1,X2)) -> proper#(X2)
p5: proper#(cons(X1,X2)) -> proper#(X1)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        proper#_A(x1) = max{4, x1 + 3}
        cons1_A(x1,x2) = max{x1, x2}
        s_A(x1) = x1 + 2
        from_A(x1) = x1
        cons_A(x1,x2) = max{x1 + 1, x2}
  
    precedence:
    
      proper# = cons1 = s = from = cons
    
    partial status:
  
      pi(proper#) = []
      pi(cons1) = [2]
      pi(s) = [1]
      pi(from) = [1]
      pi(cons) = [2]

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: proper#(cons1(X1,X2)) -> proper#(X1)
p2: proper#(from(X)) -> proper#(X)
p3: proper#(cons(X1,X2)) -> proper#(X2)
p4: proper#(cons(X1,X2)) -> proper#(X1)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: proper#(cons1(X1,X2)) -> proper#(X1)
p2: proper#(cons(X1,X2)) -> proper#(X1)
p3: proper#(cons(X1,X2)) -> proper#(X2)
p4: proper#(from(X)) -> proper#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        proper#_A(x1) = x1 + 1
        cons1_A(x1,x2) = max{x1, x2}
        cons_A(x1,x2) = max{x1, x2}
        from_A(x1) = x1
  
    precedence:
    
      proper# = cons1 = cons = from
    
    partial status:
  
      pi(proper#) = [1]
      pi(cons1) = [1, 2]
      pi(cons) = [1, 2]
      pi(from) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: proper#(cons(X1,X2)) -> proper#(X1)
p2: proper#(cons(X1,X2)) -> proper#(X2)
p3: proper#(from(X)) -> proper#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: proper#(cons(X1,X2)) -> proper#(X1)
p2: proper#(from(X)) -> proper#(X)
p3: proper#(cons(X1,X2)) -> proper#(X2)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        proper#_A(x1) = max{4, x1 + 3}
        cons_A(x1,x2) = max{x1 + 1, x2}
        from_A(x1) = x1 + 2
  
    precedence:
    
      proper# = cons = from
    
    partial status:
  
      pi(proper#) = []
      pi(cons) = [2]
      pi(from) = [1]

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: proper#(cons(X1,X2)) -> proper#(X1)
p2: proper#(cons(X1,X2)) -> proper#(X2)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: proper#(cons(X1,X2)) -> proper#(X1)
p2: proper#(cons(X1,X2)) -> proper#(X2)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        proper#_A(x1) = x1 + 3
        cons_A(x1,x2) = max{x1, x2 + 1}
  
    precedence:
    
      proper# = cons
    
    partial status:
  
      pi(proper#) = []
      pi(cons) = [2]

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: proper#(cons(X1,X2)) -> proper#(X1)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: proper#(cons(X1,X2)) -> proper#(X1)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        proper#_A(x1) = x1 + 2
        cons_A(x1,x2) = max{x1 + 1, x2 + 1}
  
    precedence:
    
      proper# = cons
    
    partial status:
  
      pi(proper#) = []
      pi(cons) = [2]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: |2nd|#(mark(X)) -> |2nd|#(X)
p2: |2nd|#(ok(X)) -> |2nd|#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        |2nd|#_A(x1) = max{3, x1 + 2}
        mark_A(x1) = x1 + 3
        ok_A(x1) = max{1, x1}
  
    precedence:
    
      |2nd|# = mark = ok
    
    partial status:
  
      pi(|2nd|#) = [1]
      pi(mark) = [1]
      pi(ok) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: |2nd|#(ok(X)) -> |2nd|#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: |2nd|#(ok(X)) -> |2nd|#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        |2nd|#_A(x1) = x1 + 2
        ok_A(x1) = x1 + 2
  
    precedence:
    
      |2nd|# = ok
    
    partial status:
  
      pi(|2nd|#) = [1]
      pi(ok) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: cons1#(mark(X1),X2) -> cons1#(X1,X2)
p2: cons1#(ok(X1),ok(X2)) -> cons1#(X1,X2)
p3: cons1#(X1,mark(X2)) -> cons1#(X1,X2)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        cons1#_A(x1,x2) = max{0, x2 - 3}
        mark_A(x1) = max{4, x1 + 2}
        ok_A(x1) = max{2, x1}
  
    precedence:
    
      cons1# = mark = ok
    
    partial status:
  
      pi(cons1#) = []
      pi(mark) = [1]
      pi(ok) = [1]

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: cons1#(mark(X1),X2) -> cons1#(X1,X2)
p2: cons1#(ok(X1),ok(X2)) -> cons1#(X1,X2)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: cons1#(mark(X1),X2) -> cons1#(X1,X2)
p2: cons1#(ok(X1),ok(X2)) -> cons1#(X1,X2)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        cons1#_A(x1,x2) = max{0, x1 - 2}
        mark_A(x1) = x1 + 3
        ok_A(x1) = max{2, x1}
  
    precedence:
    
      cons1# = mark = ok
    
    partial status:
  
      pi(cons1#) = []
      pi(mark) = [1]
      pi(ok) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: cons1#(ok(X1),ok(X2)) -> cons1#(X1,X2)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: cons1#(ok(X1),ok(X2)) -> cons1#(X1,X2)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        cons1#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
        ok_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      cons1# = ok
    
    partial status:
  
      pi(cons1#) = []
      pi(ok) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: cons#(mark(X1),X2) -> cons#(X1,X2)
p2: cons#(ok(X1),ok(X2)) -> cons#(X1,X2)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        cons#_A(x1,x2) = max{0, x1 - 2}
        mark_A(x1) = x1 + 3
        ok_A(x1) = max{2, x1}
  
    precedence:
    
      cons# = mark = ok
    
    partial status:
  
      pi(cons#) = []
      pi(mark) = [1]
      pi(ok) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: cons#(ok(X1),ok(X2)) -> cons#(X1,X2)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: cons#(ok(X1),ok(X2)) -> cons#(X1,X2)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        cons#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
        ok_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      cons# = ok
    
    partial status:
  
      pi(cons#) = []
      pi(ok) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: from#(mark(X)) -> from#(X)
p2: from#(ok(X)) -> from#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        from#_A(x1) = max{3, x1 + 2}
        mark_A(x1) = x1 + 3
        ok_A(x1) = max{1, x1}
  
    precedence:
    
      from# = mark = ok
    
    partial status:
  
      pi(from#) = [1]
      pi(mark) = [1]
      pi(ok) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: from#(ok(X)) -> from#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: from#(ok(X)) -> from#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        from#_A(x1) = x1 + 2
        ok_A(x1) = x1 + 2
  
    precedence:
    
      from# = ok
    
    partial status:
  
      pi(from#) = [1]
      pi(ok) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: s#(mark(X)) -> s#(X)
p2: s#(ok(X)) -> s#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        s#_A(x1) = max{3, x1 + 2}
        mark_A(x1) = x1
        ok_A(x1) = x1 + 1
  
    precedence:
    
      s# = mark = ok
    
    partial status:
  
      pi(s#) = [1]
      pi(mark) = [1]
      pi(ok) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: s#(ok(X)) -> s#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: s#(ok(X)) -> s#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(|2nd|(X)) -> |2nd|(active(X))
r5: active(cons(X1,X2)) -> cons(active(X1),X2)
r6: active(from(X)) -> from(active(X))
r7: active(s(X)) -> s(active(X))
r8: active(cons1(X1,X2)) -> cons1(active(X1),X2)
r9: active(cons1(X1,X2)) -> cons1(X1,active(X2))
r10: |2nd|(mark(X)) -> mark(|2nd|(X))
r11: cons(mark(X1),X2) -> mark(cons(X1,X2))
r12: from(mark(X)) -> mark(from(X))
r13: s(mark(X)) -> mark(s(X))
r14: cons1(mark(X1),X2) -> mark(cons1(X1,X2))
r15: cons1(X1,mark(X2)) -> mark(cons1(X1,X2))
r16: proper(|2nd|(X)) -> |2nd|(proper(X))
r17: proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
r18: proper(from(X)) -> from(proper(X))
r19: proper(s(X)) -> s(proper(X))
r20: proper(cons1(X1,X2)) -> cons1(proper(X1),proper(X2))
r21: |2nd|(ok(X)) -> ok(|2nd|(X))
r22: cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
r23: from(ok(X)) -> ok(from(X))
r24: s(ok(X)) -> ok(s(X))
r25: cons1(ok(X1),ok(X2)) -> ok(cons1(X1,X2))
r26: top(mark(X)) -> top(proper(X))
r27: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        s#_A(x1) = x1 + 1
        ok_A(x1) = x1
  
    precedence:
    
      s# = ok
    
    partial status:
  
      pi(s#) = [1]
      pi(ok) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.