YES

We show the termination of the TRS R:

  a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
  a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
  a__from(X) -> cons(mark(X),from(s(X)))
  mark(|2nd|(X)) -> a__2nd(mark(X))
  mark(from(X)) -> a__from(mark(X))
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(s(X)) -> s(mark(X))
  mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
  a__2nd(X) -> |2nd|(X)
  a__from(X) -> from(X)

-- SCC decomposition.

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

p1: a__2nd#(cons1(X,cons(Y,Z))) -> mark#(Y)
p2: a__2nd#(cons(X,X1)) -> a__2nd#(cons1(mark(X),mark(X1)))
p3: a__2nd#(cons(X,X1)) -> mark#(X)
p4: a__2nd#(cons(X,X1)) -> mark#(X1)
p5: a__from#(X) -> mark#(X)
p6: mark#(|2nd|(X)) -> a__2nd#(mark(X))
p7: mark#(|2nd|(X)) -> mark#(X)
p8: mark#(from(X)) -> a__from#(mark(X))
p9: mark#(from(X)) -> mark#(X)
p10: mark#(cons(X1,X2)) -> mark#(X1)
p11: mark#(s(X)) -> mark#(X)
p12: mark#(cons1(X1,X2)) -> mark#(X1)
p13: mark#(cons1(X1,X2)) -> mark#(X2)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13}


-- Reduction pair.

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

p1: a__2nd#(cons1(X,cons(Y,Z))) -> mark#(Y)
p2: mark#(cons1(X1,X2)) -> mark#(X2)
p3: mark#(cons1(X1,X2)) -> mark#(X1)
p4: mark#(s(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> mark#(X1)
p6: mark#(from(X)) -> mark#(X)
p7: mark#(from(X)) -> a__from#(mark(X))
p8: a__from#(X) -> mark#(X)
p9: mark#(|2nd|(X)) -> mark#(X)
p10: mark#(|2nd|(X)) -> a__2nd#(mark(X))
p11: a__2nd#(cons(X,X1)) -> mark#(X1)
p12: a__2nd#(cons(X,X1)) -> mark#(X)
p13: a__2nd#(cons(X,X1)) -> a__2nd#(cons1(mark(X),mark(X1)))

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__2nd#_A(x1) = ((1,1),(1,1)) x1 + (1,1)
          cons1_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(0,1)) x2 + (1,2)
          cons_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,1),(0,0)) x2 + (5,1)
          mark#_A(x1) = ((0,1),(0,1)) x1 + (1,0)
          s_A(x1) = ((0,0),(1,1)) x1 + (2,3)
          from_A(x1) = ((1,1),(1,1)) x1 + (13,2)
          a__from#_A(x1) = ((0,1),(0,1)) x1 + (2,1)
          mark_A(x1) = x1
          |2nd|_A(x1) = ((1,1),(1,1)) x1 + (3,1)
          a__2nd_A(x1) = ((1,1),(1,1)) x1 + (3,1)
          a__from_A(x1) = ((1,1),(1,1)) x1 + (13,2)
  
    precedence:
    
      a__2nd# = mark# = a__from# > cons1 = mark = a__2nd > cons = s = |2nd| = a__from > from
    
    partial status:
  
      pi(a__2nd#) = [1]
      pi(cons1) = []
      pi(cons) = []
      pi(mark#) = []
      pi(s) = []
      pi(from) = []
      pi(a__from#) = []
      pi(mark) = []
      pi(|2nd|) = []
      pi(a__2nd) = []
      pi(a__from) = []

The next rules are strictly ordered:

  p12

We remove them from the problem.

-- SCC decomposition.

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

p1: a__2nd#(cons1(X,cons(Y,Z))) -> mark#(Y)
p2: mark#(cons1(X1,X2)) -> mark#(X2)
p3: mark#(cons1(X1,X2)) -> mark#(X1)
p4: mark#(s(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> mark#(X1)
p6: mark#(from(X)) -> mark#(X)
p7: mark#(from(X)) -> a__from#(mark(X))
p8: a__from#(X) -> mark#(X)
p9: mark#(|2nd|(X)) -> mark#(X)
p10: mark#(|2nd|(X)) -> a__2nd#(mark(X))
p11: a__2nd#(cons(X,X1)) -> mark#(X1)
p12: a__2nd#(cons(X,X1)) -> a__2nd#(cons1(mark(X),mark(X1)))

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12}


-- Reduction pair.

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

p1: a__2nd#(cons1(X,cons(Y,Z))) -> mark#(Y)
p2: mark#(|2nd|(X)) -> a__2nd#(mark(X))
p3: a__2nd#(cons(X,X1)) -> a__2nd#(cons1(mark(X),mark(X1)))
p4: a__2nd#(cons(X,X1)) -> mark#(X1)
p5: mark#(|2nd|(X)) -> mark#(X)
p6: mark#(from(X)) -> a__from#(mark(X))
p7: a__from#(X) -> mark#(X)
p8: mark#(from(X)) -> mark#(X)
p9: mark#(cons(X1,X2)) -> mark#(X1)
p10: mark#(s(X)) -> mark#(X)
p11: mark#(cons1(X1,X2)) -> mark#(X1)
p12: mark#(cons1(X1,X2)) -> mark#(X2)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__2nd#_A(x1) = ((1,1),(1,1)) x1 + (1,0)
          cons1_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(0,1)) x2 + (3,1)
          cons_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,1),(0,0)) x2 + (2,2)
          mark#_A(x1) = ((0,1),(0,1)) x1 + (5,4)
          |2nd|_A(x1) = ((1,1),(1,1)) x1 + (6,1)
          mark_A(x1) = x1
          from_A(x1) = ((1,1),(1,1)) x1 + (14,2)
          a__from#_A(x1) = ((0,1),(0,1)) x1 + (6,4)
          s_A(x1) = ((0,0),(1,1)) x1 + (6,4)
          a__2nd_A(x1) = ((1,1),(1,1)) x1 + (6,1)
          a__from_A(x1) = ((1,1),(1,1)) x1 + (14,2)
  
    precedence:
    
      |2nd| = a__from# = a__2nd > cons1 = from = a__from > a__2nd# > mark# > mark = s > cons
    
    partial status:
  
      pi(a__2nd#) = []
      pi(cons1) = []
      pi(cons) = []
      pi(mark#) = []
      pi(|2nd|) = []
      pi(mark) = [1]
      pi(from) = []
      pi(a__from#) = []
      pi(s) = []
      pi(a__2nd) = []
      pi(a__from) = []

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: mark#(|2nd|(X)) -> a__2nd#(mark(X))
p2: a__2nd#(cons(X,X1)) -> a__2nd#(cons1(mark(X),mark(X1)))
p3: a__2nd#(cons(X,X1)) -> mark#(X1)
p4: mark#(|2nd|(X)) -> mark#(X)
p5: mark#(from(X)) -> a__from#(mark(X))
p6: a__from#(X) -> mark#(X)
p7: mark#(from(X)) -> mark#(X)
p8: mark#(cons(X1,X2)) -> mark#(X1)
p9: mark#(s(X)) -> mark#(X)
p10: mark#(cons1(X1,X2)) -> mark#(X1)
p11: mark#(cons1(X1,X2)) -> mark#(X2)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p3, p4, p5, p6, p7, p8, p9, p10, p11}


-- Reduction pair.

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

p1: mark#(|2nd|(X)) -> a__2nd#(mark(X))
p2: a__2nd#(cons(X,X1)) -> mark#(X1)
p3: mark#(cons1(X1,X2)) -> mark#(X2)
p4: mark#(cons1(X1,X2)) -> mark#(X1)
p5: mark#(s(X)) -> mark#(X)
p6: mark#(cons(X1,X2)) -> mark#(X1)
p7: mark#(from(X)) -> mark#(X)
p8: mark#(from(X)) -> a__from#(mark(X))
p9: a__from#(X) -> mark#(X)
p10: mark#(|2nd|(X)) -> mark#(X)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((1,0),(1,0)) x1 + (6,7)
          |2nd|_A(x1) = ((1,1),(1,1)) x1 + (2,6)
          a__2nd#_A(x1) = ((0,1),(1,1)) x1 + (1,1)
          mark_A(x1) = x1
          cons_A(x1,x2) = ((1,1),(0,0)) x1 + ((0,0),(1,0)) x2 + (5,6)
          cons1_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (1,1)
          s_A(x1) = ((1,0),(0,0)) x1 + (7,1)
          from_A(x1) = ((1,1),(1,0)) x1 + (8,22)
          a__from#_A(x1) = ((1,0),(1,0)) x1 + (7,15)
          a__2nd_A(x1) = ((1,1),(1,1)) x1 + (2,6)
          a__from_A(x1) = ((1,1),(1,0)) x1 + (8,22)
  
    precedence:
    
      a__2nd# > mark# = a__from# > mark = cons1 = a__2nd > |2nd| > a__from > cons > s > from
    
    partial status:
  
      pi(mark#) = []
      pi(|2nd|) = []
      pi(a__2nd#) = []
      pi(mark) = []
      pi(cons) = []
      pi(cons1) = []
      pi(s) = []
      pi(from) = []
      pi(a__from#) = []
      pi(a__2nd) = []
      pi(a__from) = []

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: mark#(|2nd|(X)) -> a__2nd#(mark(X))
p2: mark#(cons1(X1,X2)) -> mark#(X2)
p3: mark#(cons1(X1,X2)) -> mark#(X1)
p4: mark#(s(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> mark#(X1)
p6: mark#(from(X)) -> mark#(X)
p7: mark#(from(X)) -> a__from#(mark(X))
p8: a__from#(X) -> mark#(X)
p9: mark#(|2nd|(X)) -> mark#(X)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p5, p6, p7, p8, p9}


-- Reduction pair.

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

p1: mark#(cons1(X1,X2)) -> mark#(X2)
p2: mark#(|2nd|(X)) -> mark#(X)
p3: mark#(from(X)) -> a__from#(mark(X))
p4: a__from#(X) -> mark#(X)
p5: mark#(from(X)) -> mark#(X)
p6: mark#(cons(X1,X2)) -> mark#(X1)
p7: mark#(s(X)) -> mark#(X)
p8: mark#(cons1(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((0,1),(0,1)) x1 + (1,0)
          cons1_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,0),(0,1)) x2 + (3,1)
          |2nd|_A(x1) = ((1,1),(1,1)) x1 + (2,1)
          from_A(x1) = ((1,1),(1,1)) x1 + (10,2)
          a__from#_A(x1) = ((1,1),(0,1)) x1 + (2,0)
          mark_A(x1) = x1
          cons_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,1),(0,0)) x2 + (2,2)
          s_A(x1) = ((0,0),(1,1)) x1 + (2,3)
          a__2nd_A(x1) = ((1,1),(1,1)) x1 + (2,1)
          a__from_A(x1) = ((1,1),(1,1)) x1 + (10,2)
  
    precedence:
    
      from = mark = a__2nd = a__from > cons > mark# = |2nd| = a__from# = s > cons1
    
    partial status:
  
      pi(mark#) = []
      pi(cons1) = []
      pi(|2nd|) = []
      pi(from) = []
      pi(a__from#) = [1]
      pi(mark) = [1]
      pi(cons) = []
      pi(s) = []
      pi(a__2nd) = []
      pi(a__from) = []

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

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p4, p5, p6, p7}


-- Reduction pair.

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

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

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = x1 + (2,2)
          cons1_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (3,2)
          s_A(x1) = x1 + (1,1)
          cons_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (3,2)
          from_A(x1) = x1 + (3,2)
          |2nd|_A(x1) = ((1,1),(0,1)) x1 + (3,2)
  
    precedence:
    
      mark# > cons1 > s > cons = from = |2nd|
    
    partial status:
  
      pi(mark#) = [1]
      pi(cons1) = [1, 2]
      pi(s) = [1]
      pi(cons) = [1, 2]
      pi(from) = [1]
      pi(|2nd|) = []

The next rules are strictly ordered:

  p6

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(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: mark#(cons1(X1,X2)) -> mark#(X2)
p2: mark#(from(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(s(X)) -> mark#(X)
p5: mark#(cons1(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          cons1_A(x1,x2) = ((0,0),(0,1)) x1 + x2 + (3,2)
          from_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          cons_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
          s_A(x1) = ((0,0),(0,1)) x1 + (1,1)
  
    precedence:
    
      mark# > cons1 > from = cons = s
    
    partial status:
  
      pi(mark#) = []
      pi(cons1) = []
      pi(from) = []
      pi(cons) = []
      pi(s) = []

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: mark#(cons1(X1,X2)) -> mark#(X2)
p2: mark#(from(X)) -> mark#(X)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(cons1(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(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: mark#(cons1(X1,X2)) -> mark#(X2)
p2: mark#(cons1(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(from(X)) -> mark#(X)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          cons1_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (3,2)
          s_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          from_A(x1) = ((0,0),(0,1)) x1 + (1,1)
  
    precedence:
    
      from > mark# > cons1 = s
    
    partial status:
  
      pi(mark#) = []
      pi(cons1) = []
      pi(s) = []
      pi(from) = []

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: mark#(cons1(X1,X2)) -> mark#(X2)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(from(X)) -> mark#(X)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(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: mark#(cons1(X1,X2)) -> mark#(X2)
p2: mark#(from(X)) -> mark#(X)
p3: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          cons1_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(0,1)) x2 + (3,2)
          from_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          s_A(x1) = ((0,0),(0,1)) x1 + (1,1)
  
    precedence:
    
      mark# > cons1 = from = s
    
    partial status:
  
      pi(mark#) = []
      pi(cons1) = [2]
      pi(from) = []
      pi(s) = []

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: mark#(cons1(X1,X2)) -> mark#(X2)
p2: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(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: mark#(cons1(X1,X2)) -> mark#(X2)
p2: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((1,0),(0,0)) x1 + (2,2)
          cons1_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (1,1)
          s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      mark# = cons1 = s
    
    partial status:
  
      pi(mark#) = []
      pi(cons1) = [2]
      pi(s) = []

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: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      mark# = s
    
    partial status:
  
      pi(mark#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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