YES

We show the termination of the TRS R:

  active(incr(nil())) -> mark(nil())
  active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
  active(adx(nil())) -> mark(nil())
  active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
  active(nats()) -> mark(adx(zeros()))
  active(zeros()) -> mark(cons(|0|(),zeros()))
  active(head(cons(X,L))) -> mark(X)
  active(tail(cons(X,L))) -> mark(L)
  mark(incr(X)) -> active(incr(mark(X)))
  mark(nil()) -> active(nil())
  mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
  mark(s(X)) -> active(s(mark(X)))
  mark(adx(X)) -> active(adx(mark(X)))
  mark(nats()) -> active(nats())
  mark(zeros()) -> active(zeros())
  mark(|0|()) -> active(|0|())
  mark(head(X)) -> active(head(mark(X)))
  mark(tail(X)) -> active(tail(mark(X)))
  incr(mark(X)) -> incr(X)
  incr(active(X)) -> incr(X)
  cons(mark(X1),X2) -> cons(X1,X2)
  cons(X1,mark(X2)) -> cons(X1,X2)
  cons(active(X1),X2) -> cons(X1,X2)
  cons(X1,active(X2)) -> cons(X1,X2)
  s(mark(X)) -> s(X)
  s(active(X)) -> s(X)
  adx(mark(X)) -> adx(X)
  adx(active(X)) -> adx(X)
  head(mark(X)) -> head(X)
  head(active(X)) -> head(X)
  tail(mark(X)) -> tail(X)
  tail(active(X)) -> tail(X)

-- SCC decomposition.

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

p1: active#(incr(nil())) -> mark#(nil())
p2: active#(incr(cons(X,L))) -> mark#(cons(s(X),incr(L)))
p3: active#(incr(cons(X,L))) -> cons#(s(X),incr(L))
p4: active#(incr(cons(X,L))) -> s#(X)
p5: active#(incr(cons(X,L))) -> incr#(L)
p6: active#(adx(nil())) -> mark#(nil())
p7: active#(adx(cons(X,L))) -> mark#(incr(cons(X,adx(L))))
p8: active#(adx(cons(X,L))) -> incr#(cons(X,adx(L)))
p9: active#(adx(cons(X,L))) -> cons#(X,adx(L))
p10: active#(adx(cons(X,L))) -> adx#(L)
p11: active#(nats()) -> mark#(adx(zeros()))
p12: active#(nats()) -> adx#(zeros())
p13: active#(zeros()) -> mark#(cons(|0|(),zeros()))
p14: active#(zeros()) -> cons#(|0|(),zeros())
p15: active#(head(cons(X,L))) -> mark#(X)
p16: active#(tail(cons(X,L))) -> mark#(L)
p17: mark#(incr(X)) -> active#(incr(mark(X)))
p18: mark#(incr(X)) -> incr#(mark(X))
p19: mark#(incr(X)) -> mark#(X)
p20: mark#(nil()) -> active#(nil())
p21: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p22: mark#(cons(X1,X2)) -> cons#(mark(X1),X2)
p23: mark#(cons(X1,X2)) -> mark#(X1)
p24: mark#(s(X)) -> active#(s(mark(X)))
p25: mark#(s(X)) -> s#(mark(X))
p26: mark#(s(X)) -> mark#(X)
p27: mark#(adx(X)) -> active#(adx(mark(X)))
p28: mark#(adx(X)) -> adx#(mark(X))
p29: mark#(adx(X)) -> mark#(X)
p30: mark#(nats()) -> active#(nats())
p31: mark#(zeros()) -> active#(zeros())
p32: mark#(|0|()) -> active#(|0|())
p33: mark#(head(X)) -> active#(head(mark(X)))
p34: mark#(head(X)) -> head#(mark(X))
p35: mark#(head(X)) -> mark#(X)
p36: mark#(tail(X)) -> active#(tail(mark(X)))
p37: mark#(tail(X)) -> tail#(mark(X))
p38: mark#(tail(X)) -> mark#(X)
p39: incr#(mark(X)) -> incr#(X)
p40: incr#(active(X)) -> incr#(X)
p41: cons#(mark(X1),X2) -> cons#(X1,X2)
p42: cons#(X1,mark(X2)) -> cons#(X1,X2)
p43: cons#(active(X1),X2) -> cons#(X1,X2)
p44: cons#(X1,active(X2)) -> cons#(X1,X2)
p45: s#(mark(X)) -> s#(X)
p46: s#(active(X)) -> s#(X)
p47: adx#(mark(X)) -> adx#(X)
p48: adx#(active(X)) -> adx#(X)
p49: head#(mark(X)) -> head#(X)
p50: head#(active(X)) -> head#(X)
p51: tail#(mark(X)) -> tail#(X)
p52: tail#(active(X)) -> tail#(X)

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(X)

The estimated dependency graph contains the following SCCs:

  {p2, p7, p11, p13, p15, p16, p17, p19, p21, p23, p24, p26, p27, p29, p30, p31, p33, p35, p36, p38}
  {p41, p42, p43, p44}
  {p45, p46}
  {p39, p40}
  {p47, p48}
  {p49, p50}
  {p51, p52}


-- Reduction pair.

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

p1: mark#(tail(X)) -> active#(tail(mark(X)))
p2: active#(tail(cons(X,L))) -> mark#(L)
p3: mark#(tail(X)) -> mark#(X)
p4: mark#(head(X)) -> mark#(X)
p5: mark#(head(X)) -> active#(head(mark(X)))
p6: active#(head(cons(X,L))) -> mark#(X)
p7: mark#(zeros()) -> active#(zeros())
p8: active#(zeros()) -> mark#(cons(|0|(),zeros()))
p9: mark#(cons(X1,X2)) -> mark#(X1)
p10: mark#(nats()) -> active#(nats())
p11: active#(nats()) -> mark#(adx(zeros()))
p12: mark#(adx(X)) -> mark#(X)
p13: mark#(adx(X)) -> active#(adx(mark(X)))
p14: active#(adx(cons(X,L))) -> mark#(incr(cons(X,adx(L))))
p15: mark#(s(X)) -> mark#(X)
p16: mark#(s(X)) -> active#(s(mark(X)))
p17: active#(incr(cons(X,L))) -> mark#(cons(s(X),incr(L)))
p18: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p19: mark#(incr(X)) -> mark#(X)
p20: mark#(incr(X)) -> active#(incr(mark(X)))

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(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, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{7, x1 + 5}
          tail_A(x1) = max{61, x1 + 37}
          active#_A(x1) = max{25, x1 + 5}
          mark_A(x1) = max{23, x1}
          cons_A(x1,x2) = max{30, x1 + 6, x2 - 13}
          head_A(x1) = max{23, x1}
          zeros_A = 35
          |0|_A = 22
          nats_A = 70
          adx_A(x1) = max{70, x1 + 21}
          incr_A(x1) = max{23, x1}
          s_A(x1) = max{23, x1}
          active_A(x1) = max{23, x1}
          nil_A = 71
    
      precedence:
      
        tail = mark = head = zeros = |0| = nats = adx = incr = s = active > mark# = active# > cons = nil
      
      partial status:
    
        pi(mark#) = [1]
        pi(tail) = []
        pi(active#) = [1]
        pi(mark) = []
        pi(cons) = []
        pi(head) = []
        pi(zeros) = []
        pi(|0|) = []
        pi(nats) = []
        pi(adx) = []
        pi(incr) = []
        pi(s) = []
        pi(active) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 265
          tail_A(x1) = 278
          active#_A(x1) = 265
          mark_A(x1) = 266
          cons_A(x1,x2) = 258
          head_A(x1) = 278
          zeros_A = 149
          |0|_A = 278
          nats_A = 0
          adx_A(x1) = 235
          incr_A(x1) = 265
          s_A(x1) = 267
          active_A(x1) = 266
          nil_A = 307
    
      precedence:
      
        nil > incr > mark# = tail = active# = mark = cons = head = zeros = |0| = nats = s = active > adx
      
      partial status:
    
        pi(mark#) = []
        pi(tail) = []
        pi(active#) = []
        pi(mark) = []
        pi(cons) = []
        pi(head) = []
        pi(zeros) = []
        pi(|0|) = []
        pi(nats) = []
        pi(adx) = []
        pi(incr) = []
        pi(s) = []
        pi(active) = []
        pi(nil) = []
    

The next rules are strictly ordered:

  p2, p6

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(tail(X)) -> active#(tail(mark(X)))
p2: mark#(tail(X)) -> mark#(X)
p3: mark#(head(X)) -> mark#(X)
p4: mark#(head(X)) -> active#(head(mark(X)))
p5: mark#(zeros()) -> active#(zeros())
p6: active#(zeros()) -> mark#(cons(|0|(),zeros()))
p7: mark#(cons(X1,X2)) -> mark#(X1)
p8: mark#(nats()) -> active#(nats())
p9: active#(nats()) -> mark#(adx(zeros()))
p10: mark#(adx(X)) -> mark#(X)
p11: mark#(adx(X)) -> active#(adx(mark(X)))
p12: active#(adx(cons(X,L))) -> mark#(incr(cons(X,adx(L))))
p13: mark#(s(X)) -> mark#(X)
p14: mark#(s(X)) -> active#(s(mark(X)))
p15: active#(incr(cons(X,L))) -> mark#(cons(s(X),incr(L)))
p16: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p17: mark#(incr(X)) -> mark#(X)
p18: mark#(incr(X)) -> active#(incr(mark(X)))

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p18}


-- Reduction pair.

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

p1: mark#(tail(X)) -> active#(tail(mark(X)))
p2: active#(incr(cons(X,L))) -> mark#(cons(s(X),incr(L)))
p3: mark#(incr(X)) -> active#(incr(mark(X)))
p4: active#(adx(cons(X,L))) -> mark#(incr(cons(X,adx(L))))
p5: mark#(incr(X)) -> mark#(X)
p6: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p7: active#(nats()) -> mark#(adx(zeros()))
p8: mark#(adx(X)) -> active#(adx(mark(X)))
p9: active#(zeros()) -> mark#(cons(|0|(),zeros()))
p10: mark#(cons(X1,X2)) -> mark#(X1)
p11: mark#(s(X)) -> active#(s(mark(X)))
p12: mark#(s(X)) -> mark#(X)
p13: mark#(adx(X)) -> mark#(X)
p14: mark#(nats()) -> active#(nats())
p15: mark#(zeros()) -> active#(zeros())
p16: mark#(head(X)) -> active#(head(mark(X)))
p17: mark#(head(X)) -> mark#(X)
p18: mark#(tail(X)) -> mark#(X)

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(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, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 4
          tail_A(x1) = x1 + 26
          active#_A(x1) = x1 + 4
          mark_A(x1) = x1
          incr_A(x1) = max{49, x1}
          cons_A(x1,x2) = max{18, x1 + 6, x2 - 25}
          s_A(x1) = max{7, x1}
          adx_A(x1) = x1 + 48
          nats_A = 76
          zeros_A = 23
          |0|_A = 2
          head_A(x1) = max{31, x1 + 26}
          active_A(x1) = max{1, x1}
          nil_A = 2
    
      precedence:
      
        incr = s = adx > zeros > cons > mark# = active# = nats > |0| > tail = mark = head = active = nil
      
      partial status:
    
        pi(mark#) = [1]
        pi(tail) = []
        pi(active#) = [1]
        pi(mark) = []
        pi(incr) = []
        pi(cons) = []
        pi(s) = []
        pi(adx) = []
        pi(nats) = []
        pi(zeros) = []
        pi(|0|) = []
        pi(head) = []
        pi(active) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 49
          tail_A(x1) = 45
          active#_A(x1) = max{49, x1 - 3}
          mark_A(x1) = 167
          incr_A(x1) = 49
          cons_A(x1,x2) = 43
          s_A(x1) = 45
          adx_A(x1) = 51
          nats_A = 48
          zeros_A = 52
          |0|_A = 0
          head_A(x1) = 45
          active_A(x1) = 167
          nil_A = 0
    
      precedence:
      
        tail = mark = cons = s = nats = zeros = head = active = nil > mark# = active# = incr = |0| > adx
      
      partial status:
    
        pi(mark#) = []
        pi(tail) = []
        pi(active#) = []
        pi(mark) = []
        pi(incr) = []
        pi(cons) = []
        pi(s) = []
        pi(adx) = []
        pi(nats) = []
        pi(zeros) = []
        pi(|0|) = []
        pi(head) = []
        pi(active) = []
        pi(nil) = []
    

The next rules are strictly ordered:

  p2, p7

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(tail(X)) -> active#(tail(mark(X)))
p2: mark#(incr(X)) -> active#(incr(mark(X)))
p3: active#(adx(cons(X,L))) -> mark#(incr(cons(X,adx(L))))
p4: mark#(incr(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p6: mark#(adx(X)) -> active#(adx(mark(X)))
p7: active#(zeros()) -> mark#(cons(|0|(),zeros()))
p8: mark#(cons(X1,X2)) -> mark#(X1)
p9: mark#(s(X)) -> active#(s(mark(X)))
p10: mark#(s(X)) -> mark#(X)
p11: mark#(adx(X)) -> mark#(X)
p12: mark#(nats()) -> active#(nats())
p13: mark#(zeros()) -> active#(zeros())
p14: mark#(head(X)) -> active#(head(mark(X)))
p15: mark#(head(X)) -> mark#(X)
p16: mark#(tail(X)) -> mark#(X)

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p13, p14, p15, p16}


-- Reduction pair.

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

p1: mark#(tail(X)) -> active#(tail(mark(X)))
p2: active#(zeros()) -> mark#(cons(|0|(),zeros()))
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(tail(X)) -> mark#(X)
p5: mark#(head(X)) -> mark#(X)
p6: mark#(head(X)) -> active#(head(mark(X)))
p7: active#(adx(cons(X,L))) -> mark#(incr(cons(X,adx(L))))
p8: mark#(zeros()) -> active#(zeros())
p9: mark#(adx(X)) -> mark#(X)
p10: mark#(s(X)) -> mark#(X)
p11: mark#(s(X)) -> active#(s(mark(X)))
p12: mark#(adx(X)) -> active#(adx(mark(X)))
p13: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p14: mark#(incr(X)) -> mark#(X)
p15: mark#(incr(X)) -> active#(incr(mark(X)))

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(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, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{1, x1 - 2}
          tail_A(x1) = x1 + 14
          active#_A(x1) = max{1, x1 - 2}
          mark_A(x1) = x1
          zeros_A = 12
          cons_A(x1,x2) = max{10, x1 + 4, x2 - 13}
          |0|_A = 6
          head_A(x1) = x1 + 14
          adx_A(x1) = x1 + 1
          incr_A(x1) = max{1, x1}
          s_A(x1) = max{5, x1}
          active_A(x1) = x1
          nil_A = 2
          nats_A = 14
    
      precedence:
      
        tail = head > s > zeros = |0| > mark# = active# = mark = cons = adx = incr = active = nil = nats
      
      partial status:
    
        pi(mark#) = []
        pi(tail) = []
        pi(active#) = []
        pi(mark) = []
        pi(zeros) = []
        pi(cons) = []
        pi(|0|) = []
        pi(head) = []
        pi(adx) = []
        pi(incr) = []
        pi(s) = []
        pi(active) = []
        pi(nil) = []
        pi(nats) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 75
          tail_A(x1) = 130
          active#_A(x1) = 75
          mark_A(x1) = 127
          zeros_A = 129
          cons_A(x1,x2) = 129
          |0|_A = 135
          head_A(x1) = 133
          adx_A(x1) = 127
          incr_A(x1) = 74
          s_A(x1) = 131
          active_A(x1) = 127
          nil_A = 138
          nats_A = 116
    
      precedence:
      
        s > |0| > mark = head = adx = active = nil = nats > mark# = active# = zeros > cons > tail = incr
      
      partial status:
    
        pi(mark#) = []
        pi(tail) = []
        pi(active#) = []
        pi(mark) = []
        pi(zeros) = []
        pi(cons) = []
        pi(|0|) = []
        pi(head) = []
        pi(adx) = []
        pi(incr) = []
        pi(s) = []
        pi(active) = []
        pi(nil) = []
        pi(nats) = []
    

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#(tail(X)) -> active#(tail(mark(X)))
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(tail(X)) -> mark#(X)
p4: mark#(head(X)) -> mark#(X)
p5: mark#(head(X)) -> active#(head(mark(X)))
p6: active#(adx(cons(X,L))) -> mark#(incr(cons(X,adx(L))))
p7: mark#(zeros()) -> active#(zeros())
p8: mark#(adx(X)) -> mark#(X)
p9: mark#(s(X)) -> mark#(X)
p10: mark#(s(X)) -> active#(s(mark(X)))
p11: mark#(adx(X)) -> active#(adx(mark(X)))
p12: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p13: mark#(incr(X)) -> mark#(X)
p14: mark#(incr(X)) -> active#(incr(mark(X)))

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p8, p9, p10, p11, p12, p13, p14}


-- Reduction pair.

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

p1: mark#(tail(X)) -> active#(tail(mark(X)))
p2: active#(adx(cons(X,L))) -> mark#(incr(cons(X,adx(L))))
p3: mark#(incr(X)) -> active#(incr(mark(X)))
p4: mark#(incr(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p6: mark#(adx(X)) -> active#(adx(mark(X)))
p7: mark#(s(X)) -> active#(s(mark(X)))
p8: mark#(s(X)) -> mark#(X)
p9: mark#(adx(X)) -> mark#(X)
p10: mark#(head(X)) -> active#(head(mark(X)))
p11: mark#(head(X)) -> mark#(X)
p12: mark#(tail(X)) -> mark#(X)
p13: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(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, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 43
          tail_A(x1) = max{51, x1 + 45}
          active#_A(x1) = max{44, x1 + 43}
          mark_A(x1) = max{3, x1}
          adx_A(x1) = max{132, x1 + 85}
          cons_A(x1,x2) = max{16, x1 + 8, x2 - 12}
          incr_A(x1) = max{29, x1}
          s_A(x1) = max{7, x1}
          head_A(x1) = max{6, x1 + 2}
          active_A(x1) = max{6, x1}
          nil_A = 7
          nats_A = 133
          zeros_A = 17
          |0|_A = 9
    
      precedence:
      
        mark# = active# = |0| > adx > incr > zeros > cons > tail = mark > active > head = nil = nats > s
      
      partial status:
    
        pi(mark#) = []
        pi(tail) = [1]
        pi(active#) = []
        pi(mark) = [1]
        pi(adx) = []
        pi(cons) = []
        pi(incr) = []
        pi(s) = []
        pi(head) = [1]
        pi(active) = [1]
        pi(nil) = []
        pi(nats) = []
        pi(zeros) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 82
          tail_A(x1) = 102
          active#_A(x1) = 82
          mark_A(x1) = 101
          adx_A(x1) = 82
          cons_A(x1,x2) = 125
          incr_A(x1) = 101
          s_A(x1) = 101
          head_A(x1) = 102
          active_A(x1) = 101
          nil_A = 100
          nats_A = 85
          zeros_A = 103
          |0|_A = 102
    
      precedence:
      
        incr = head > mark# = tail = active# = mark = adx = cons = s = active = nil = nats = zeros = |0|
      
      partial status:
    
        pi(mark#) = []
        pi(tail) = []
        pi(active#) = []
        pi(mark) = []
        pi(adx) = []
        pi(cons) = []
        pi(incr) = []
        pi(s) = []
        pi(head) = []
        pi(active) = []
        pi(nil) = []
        pi(nats) = []
        pi(zeros) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p9, p13

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(tail(X)) -> active#(tail(mark(X)))
p2: active#(adx(cons(X,L))) -> mark#(incr(cons(X,adx(L))))
p3: mark#(incr(X)) -> active#(incr(mark(X)))
p4: mark#(incr(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p6: mark#(adx(X)) -> active#(adx(mark(X)))
p7: mark#(s(X)) -> active#(s(mark(X)))
p8: mark#(s(X)) -> mark#(X)
p9: mark#(head(X)) -> active#(head(mark(X)))
p10: mark#(head(X)) -> mark#(X)
p11: mark#(tail(X)) -> mark#(X)

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: mark#(tail(X)) -> active#(tail(mark(X)))
p2: active#(adx(cons(X,L))) -> mark#(incr(cons(X,adx(L))))
p3: mark#(tail(X)) -> mark#(X)
p4: mark#(head(X)) -> mark#(X)
p5: mark#(head(X)) -> active#(head(mark(X)))
p6: mark#(s(X)) -> mark#(X)
p7: mark#(s(X)) -> active#(s(mark(X)))
p8: mark#(adx(X)) -> active#(adx(mark(X)))
p9: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p10: mark#(incr(X)) -> mark#(X)
p11: mark#(incr(X)) -> active#(incr(mark(X)))

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(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, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{12, x1 + 6}
          tail_A(x1) = max{2, x1 + 1}
          active#_A(x1) = x1 + 6
          mark_A(x1) = x1
          adx_A(x1) = x1 + 21
          cons_A(x1,x2) = max{13, x1 + 5, x2}
          incr_A(x1) = max{4, x1}
          head_A(x1) = x1 + 8
          s_A(x1) = max{5, x1}
          active_A(x1) = max{2, x1}
          nil_A = 3
          nats_A = 35
          zeros_A = 13
          |0|_A = 8
    
      precedence:
      
        mark = incr = head = active = nats = zeros > mark# = tail = active# = s > adx = nil > cons = |0|
      
      partial status:
    
        pi(mark#) = []
        pi(tail) = []
        pi(active#) = []
        pi(mark) = []
        pi(adx) = [1]
        pi(cons) = []
        pi(incr) = []
        pi(head) = []
        pi(s) = []
        pi(active) = []
        pi(nil) = []
        pi(nats) = []
        pi(zeros) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 88
          tail_A(x1) = 52
          active#_A(x1) = 88
          mark_A(x1) = 43
          adx_A(x1) = 56
          cons_A(x1,x2) = 32
          incr_A(x1) = 54
          head_A(x1) = 89
          s_A(x1) = 88
          active_A(x1) = 43
          nil_A = 80
          nats_A = 0
          zeros_A = 2
          |0|_A = 52
    
      precedence:
      
        nil = nats = |0| > mark# = tail = active# = mark = adx = cons = incr = head = active = zeros > s
      
      partial status:
    
        pi(mark#) = []
        pi(tail) = []
        pi(active#) = []
        pi(mark) = []
        pi(adx) = []
        pi(cons) = []
        pi(incr) = []
        pi(head) = []
        pi(s) = []
        pi(active) = []
        pi(nil) = []
        pi(nats) = []
        pi(zeros) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(tail(X)) -> active#(tail(mark(X)))
p2: active#(adx(cons(X,L))) -> mark#(incr(cons(X,adx(L))))
p3: mark#(tail(X)) -> mark#(X)
p4: mark#(head(X)) -> active#(head(mark(X)))
p5: mark#(s(X)) -> mark#(X)
p6: mark#(s(X)) -> active#(s(mark(X)))
p7: mark#(adx(X)) -> active#(adx(mark(X)))
p8: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p9: mark#(incr(X)) -> mark#(X)
p10: mark#(incr(X)) -> active#(incr(mark(X)))

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: mark#(tail(X)) -> active#(tail(mark(X)))
p2: active#(adx(cons(X,L))) -> mark#(incr(cons(X,adx(L))))
p3: mark#(incr(X)) -> active#(incr(mark(X)))
p4: mark#(incr(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p6: mark#(adx(X)) -> active#(adx(mark(X)))
p7: mark#(s(X)) -> active#(s(mark(X)))
p8: mark#(s(X)) -> mark#(X)
p9: mark#(head(X)) -> active#(head(mark(X)))
p10: mark#(tail(X)) -> mark#(X)

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(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, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{9, x1 - 9}
          tail_A(x1) = max{21, x1 + 19}
          active#_A(x1) = max{1, x1 - 9}
          mark_A(x1) = max{2, x1}
          adx_A(x1) = max{24, x1}
          cons_A(x1,x2) = max{18, x1 + 6, x2}
          incr_A(x1) = max{24, x1}
          s_A(x1) = max{18, x1}
          head_A(x1) = max{0, x1 - 3}
          active_A(x1) = x1
          nil_A = 1
          nats_A = 25
          zeros_A = 19
          |0|_A = 1
    
      precedence:
      
        zeros > mark# = tail = active# > mark = adx = cons = incr = active = nil > s = head = nats = |0|
      
      partial status:
    
        pi(mark#) = []
        pi(tail) = []
        pi(active#) = []
        pi(mark) = [1]
        pi(adx) = []
        pi(cons) = []
        pi(incr) = []
        pi(s) = []
        pi(head) = []
        pi(active) = [1]
        pi(nil) = []
        pi(nats) = []
        pi(zeros) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 195
          tail_A(x1) = 88
          active#_A(x1) = 195
          mark_A(x1) = max{2, x1 - 22}
          adx_A(x1) = 101
          cons_A(x1,x2) = 93
          incr_A(x1) = 97
          s_A(x1) = 87
          head_A(x1) = 87
          active_A(x1) = max{65, x1 - 25}
          nil_A = 28
          nats_A = 105
          zeros_A = 106
          |0|_A = 86
    
      precedence:
      
        adx = nats = |0| > nil > mark# = active# = mark = cons = incr = s = zeros > tail = head = active
      
      partial status:
    
        pi(mark#) = []
        pi(tail) = []
        pi(active#) = []
        pi(mark) = []
        pi(adx) = []
        pi(cons) = []
        pi(incr) = []
        pi(s) = []
        pi(head) = []
        pi(active) = []
        pi(nil) = []
        pi(nats) = []
        pi(zeros) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p10

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(tail(X)) -> active#(tail(mark(X)))
p2: active#(adx(cons(X,L))) -> mark#(incr(cons(X,adx(L))))
p3: mark#(incr(X)) -> active#(incr(mark(X)))
p4: mark#(incr(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p6: mark#(adx(X)) -> active#(adx(mark(X)))
p7: mark#(s(X)) -> active#(s(mark(X)))
p8: mark#(s(X)) -> mark#(X)
p9: mark#(head(X)) -> active#(head(mark(X)))

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: mark#(tail(X)) -> active#(tail(mark(X)))
p2: active#(adx(cons(X,L))) -> mark#(incr(cons(X,adx(L))))
p3: mark#(head(X)) -> active#(head(mark(X)))
p4: mark#(s(X)) -> mark#(X)
p5: mark#(s(X)) -> active#(s(mark(X)))
p6: mark#(adx(X)) -> active#(adx(mark(X)))
p7: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p8: mark#(incr(X)) -> mark#(X)
p9: mark#(incr(X)) -> active#(incr(mark(X)))

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(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, r26, r27, r28, r29, r30, r31, r32

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 1
          tail_A(x1) = x1 + 4
          active#_A(x1) = x1 + 1
          mark_A(x1) = x1
          adx_A(x1) = x1 + 5
          cons_A(x1,x2) = max{x1, x2 - 3}
          incr_A(x1) = x1
          head_A(x1) = x1 + 4
          s_A(x1) = x1
          active_A(x1) = x1
          nil_A = 1
          nats_A = 8
          zeros_A = 2
          |0|_A = 1
    
      precedence:
      
        mark# = tail = active# = mark = adx = cons = incr = head = s = active = nil = nats = zeros = |0|
      
      partial status:
    
        pi(mark#) = []
        pi(tail) = []
        pi(active#) = []
        pi(mark) = []
        pi(adx) = []
        pi(cons) = []
        pi(incr) = []
        pi(head) = []
        pi(s) = []
        pi(active) = []
        pi(nil) = []
        pi(nats) = []
        pi(zeros) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 12
          tail_A(x1) = x1 + 12
          active#_A(x1) = max{1, x1}
          mark_A(x1) = x1 + 9
          adx_A(x1) = x1 + 21
          cons_A(x1,x2) = 16
          incr_A(x1) = max{15, x1 + 9}
          head_A(x1) = max{0, x1 - 10}
          s_A(x1) = max{14, x1 + 13}
          active_A(x1) = x1
          nil_A = 16
          nats_A = 27
          zeros_A = 26
          |0|_A = 27
    
      precedence:
      
        cons > |0| > nil > adx > tail = mark = active > mark# = active# = incr = head = s = nats = zeros
      
      partial status:
    
        pi(mark#) = [1]
        pi(tail) = []
        pi(active#) = [1]
        pi(mark) = []
        pi(adx) = []
        pi(cons) = []
        pi(incr) = []
        pi(head) = []
        pi(s) = []
        pi(active) = []
        pi(nil) = []
        pi(nats) = []
        pi(zeros) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p2, p4, p8

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(tail(X)) -> active#(tail(mark(X)))
p2: mark#(head(X)) -> active#(head(mark(X)))
p3: mark#(s(X)) -> active#(s(mark(X)))
p4: mark#(adx(X)) -> active#(adx(mark(X)))
p5: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p6: mark#(incr(X)) -> active#(incr(mark(X)))

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(X)

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

p1: cons#(mark(X1),X2) -> cons#(X1,X2)
p2: cons#(X1,active(X2)) -> cons#(X1,X2)
p3: cons#(active(X1),X2) -> cons#(X1,X2)
p4: cons#(X1,mark(X2)) -> cons#(X1,X2)

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          cons#_A(x1,x2) = max{2, x1 + 1, x2 + 1}
          mark_A(x1) = max{1, x1}
          active_A(x1) = max{1, x1}
    
      precedence:
      
        cons# = mark = active
      
      partial status:
    
        pi(cons#) = [1, 2]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          cons#_A(x1,x2) = max{x1 + 1, x2 - 1}
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        cons# = mark = active
      
      partial status:
    
        pi(cons#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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#(active(X)) -> s#(X)

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          s#_A(x1) = max{6, x1 + 4}
          mark_A(x1) = max{4, x1 + 3}
          active_A(x1) = max{2, x1 + 1}
    
      precedence:
      
        s# = mark = active
      
      partial status:
    
        pi(s#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          s#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        s# = mark = active
      
      partial status:
    
        pi(s#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2

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: incr#(mark(X)) -> incr#(X)
p2: incr#(active(X)) -> incr#(X)

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          incr#_A(x1) = x1 + 2
          mark_A(x1) = x1
          active_A(x1) = x1 + 2
    
      precedence:
      
        mark > active > incr#
      
      partial status:
    
        pi(incr#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          incr#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        incr# = mark = active
      
      partial status:
    
        pi(incr#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2

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: adx#(mark(X)) -> adx#(X)
p2: adx#(active(X)) -> adx#(X)

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          adx#_A(x1) = x1 + 2
          mark_A(x1) = x1
          active_A(x1) = x1 + 2
    
      precedence:
      
        mark > active > adx#
      
      partial status:
    
        pi(adx#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          adx#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        adx# = mark = active
      
      partial status:
    
        pi(adx#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2

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: head#(mark(X)) -> head#(X)
p2: head#(active(X)) -> head#(X)

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          head#_A(x1) = x1 + 2
          mark_A(x1) = x1
          active_A(x1) = x1 + 2
    
      precedence:
      
        mark > active > head#
      
      partial status:
    
        pi(head#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          head#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        head# = mark = active
      
      partial status:
    
        pi(head#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2

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: tail#(mark(X)) -> tail#(X)
p2: tail#(active(X)) -> tail#(X)

and R consists of:

r1: active(incr(nil())) -> mark(nil())
r2: active(incr(cons(X,L))) -> mark(cons(s(X),incr(L)))
r3: active(adx(nil())) -> mark(nil())
r4: active(adx(cons(X,L))) -> mark(incr(cons(X,adx(L))))
r5: active(nats()) -> mark(adx(zeros()))
r6: active(zeros()) -> mark(cons(|0|(),zeros()))
r7: active(head(cons(X,L))) -> mark(X)
r8: active(tail(cons(X,L))) -> mark(L)
r9: mark(incr(X)) -> active(incr(mark(X)))
r10: mark(nil()) -> active(nil())
r11: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(adx(X)) -> active(adx(mark(X)))
r14: mark(nats()) -> active(nats())
r15: mark(zeros()) -> active(zeros())
r16: mark(|0|()) -> active(|0|())
r17: mark(head(X)) -> active(head(mark(X)))
r18: mark(tail(X)) -> active(tail(mark(X)))
r19: incr(mark(X)) -> incr(X)
r20: incr(active(X)) -> incr(X)
r21: cons(mark(X1),X2) -> cons(X1,X2)
r22: cons(X1,mark(X2)) -> cons(X1,X2)
r23: cons(active(X1),X2) -> cons(X1,X2)
r24: cons(X1,active(X2)) -> cons(X1,X2)
r25: s(mark(X)) -> s(X)
r26: s(active(X)) -> s(X)
r27: adx(mark(X)) -> adx(X)
r28: adx(active(X)) -> adx(X)
r29: head(mark(X)) -> head(X)
r30: head(active(X)) -> head(X)
r31: tail(mark(X)) -> tail(X)
r32: tail(active(X)) -> tail(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          tail#_A(x1) = max{6, x1 + 4}
          mark_A(x1) = max{4, x1 + 3}
          active_A(x1) = max{2, x1 + 1}
    
      precedence:
      
        tail# = mark = active
      
      partial status:
    
        pi(tail#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          tail#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        tail# = mark = active
      
      partial status:
    
        pi(tail#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2

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