YES

We show the termination of the TRS R:

  a__incr(nil()) -> nil()
  a__incr(cons(X,L)) -> cons(s(mark(X)),incr(L))
  a__adx(nil()) -> nil()
  a__adx(cons(X,L)) -> a__incr(cons(mark(X),adx(L)))
  a__nats() -> a__adx(a__zeros())
  a__zeros() -> cons(|0|(),zeros())
  a__head(cons(X,L)) -> mark(X)
  a__tail(cons(X,L)) -> mark(L)
  mark(incr(X)) -> a__incr(mark(X))
  mark(adx(X)) -> a__adx(mark(X))
  mark(nats()) -> a__nats()
  mark(zeros()) -> a__zeros()
  mark(head(X)) -> a__head(mark(X))
  mark(tail(X)) -> a__tail(mark(X))
  mark(nil()) -> nil()
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(s(X)) -> s(mark(X))
  mark(|0|()) -> |0|()
  a__incr(X) -> incr(X)
  a__adx(X) -> adx(X)
  a__nats() -> nats()
  a__zeros() -> zeros()
  a__head(X) -> head(X)
  a__tail(X) -> tail(X)

-- SCC decomposition.

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

p1: a__incr#(cons(X,L)) -> mark#(X)
p2: a__adx#(cons(X,L)) -> a__incr#(cons(mark(X),adx(L)))
p3: a__adx#(cons(X,L)) -> mark#(X)
p4: a__nats#() -> a__adx#(a__zeros())
p5: a__nats#() -> a__zeros#()
p6: a__head#(cons(X,L)) -> mark#(X)
p7: a__tail#(cons(X,L)) -> mark#(L)
p8: mark#(incr(X)) -> a__incr#(mark(X))
p9: mark#(incr(X)) -> mark#(X)
p10: mark#(adx(X)) -> a__adx#(mark(X))
p11: mark#(adx(X)) -> mark#(X)
p12: mark#(nats()) -> a__nats#()
p13: mark#(zeros()) -> a__zeros#()
p14: mark#(head(X)) -> a__head#(mark(X))
p15: mark#(head(X)) -> mark#(X)
p16: mark#(tail(X)) -> a__tail#(mark(X))
p17: mark#(tail(X)) -> mark#(X)
p18: mark#(cons(X1,X2)) -> mark#(X1)
p19: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__incr(nil()) -> nil()
r2: a__incr(cons(X,L)) -> cons(s(mark(X)),incr(L))
r3: a__adx(nil()) -> nil()
r4: a__adx(cons(X,L)) -> a__incr(cons(mark(X),adx(L)))
r5: a__nats() -> a__adx(a__zeros())
r6: a__zeros() -> cons(|0|(),zeros())
r7: a__head(cons(X,L)) -> mark(X)
r8: a__tail(cons(X,L)) -> mark(L)
r9: mark(incr(X)) -> a__incr(mark(X))
r10: mark(adx(X)) -> a__adx(mark(X))
r11: mark(nats()) -> a__nats()
r12: mark(zeros()) -> a__zeros()
r13: mark(head(X)) -> a__head(mark(X))
r14: mark(tail(X)) -> a__tail(mark(X))
r15: mark(nil()) -> nil()
r16: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r17: mark(s(X)) -> s(mark(X))
r18: mark(|0|()) -> |0|()
r19: a__incr(X) -> incr(X)
r20: a__adx(X) -> adx(X)
r21: a__nats() -> nats()
r22: a__zeros() -> zeros()
r23: a__head(X) -> head(X)
r24: a__tail(X) -> tail(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p6, p7, p8, p9, p10, p11, p12, p14, p15, p16, p17, p18, p19}


-- Reduction pair.

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

p1: a__incr#(cons(X,L)) -> mark#(X)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(tail(X)) -> mark#(X)
p5: mark#(tail(X)) -> a__tail#(mark(X))
p6: a__tail#(cons(X,L)) -> mark#(L)
p7: mark#(head(X)) -> mark#(X)
p8: mark#(head(X)) -> a__head#(mark(X))
p9: a__head#(cons(X,L)) -> mark#(X)
p10: mark#(nats()) -> a__nats#()
p11: a__nats#() -> a__adx#(a__zeros())
p12: a__adx#(cons(X,L)) -> mark#(X)
p13: mark#(adx(X)) -> mark#(X)
p14: mark#(adx(X)) -> a__adx#(mark(X))
p15: a__adx#(cons(X,L)) -> a__incr#(cons(mark(X),adx(L)))
p16: mark#(incr(X)) -> mark#(X)
p17: mark#(incr(X)) -> a__incr#(mark(X))

and R consists of:

r1: a__incr(nil()) -> nil()
r2: a__incr(cons(X,L)) -> cons(s(mark(X)),incr(L))
r3: a__adx(nil()) -> nil()
r4: a__adx(cons(X,L)) -> a__incr(cons(mark(X),adx(L)))
r5: a__nats() -> a__adx(a__zeros())
r6: a__zeros() -> cons(|0|(),zeros())
r7: a__head(cons(X,L)) -> mark(X)
r8: a__tail(cons(X,L)) -> mark(L)
r9: mark(incr(X)) -> a__incr(mark(X))
r10: mark(adx(X)) -> a__adx(mark(X))
r11: mark(nats()) -> a__nats()
r12: mark(zeros()) -> a__zeros()
r13: mark(head(X)) -> a__head(mark(X))
r14: mark(tail(X)) -> a__tail(mark(X))
r15: mark(nil()) -> nil()
r16: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r17: mark(s(X)) -> s(mark(X))
r18: mark(|0|()) -> |0|()
r19: a__incr(X) -> incr(X)
r20: a__adx(X) -> adx(X)
r21: a__nats() -> nats()
r22: a__zeros() -> zeros()
r23: a__head(X) -> head(X)
r24: a__tail(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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__incr#_A(x1) = max{7, x1}
          cons_A(x1,x2) = max{5, x1, x2 - 2}
          mark#_A(x1) = max{7, x1}
          s_A(x1) = max{7, x1}
          tail_A(x1) = max{17, x1 + 10}
          a__tail#_A(x1) = max{3, x1 + 2}
          mark_A(x1) = max{7, x1}
          head_A(x1) = max{25, x1 + 18}
          a__head#_A(x1) = max{25, x1 + 10}
          nats_A = 11
          a__nats#_A = 10
          a__adx#_A(x1) = max{4, x1 + 3}
          a__zeros_A = 6
          adx_A(x1) = max{10, x1 + 3}
          incr_A(x1) = max{1, x1}
          a__incr_A(x1) = max{7, x1}
          nil_A = 8
          a__adx_A(x1) = max{10, x1 + 3}
          a__nats_A = 11
          a__head_A(x1) = max{25, x1 + 18}
          a__tail_A(x1) = max{17, x1 + 10}
          |0|_A = 6
          zeros_A = 5
    
      precedence:
      
        |0| > head = a__head > a__tail# > a__incr# = mark# = s = a__head# = a__nats# = a__adx# > mark = a__nats > adx = a__adx > a__zeros > cons = a__incr = nil > nats > tail = a__tail > incr = zeros
      
      partial status:
    
        pi(a__incr#) = []
        pi(cons) = []
        pi(mark#) = []
        pi(s) = []
        pi(tail) = []
        pi(a__tail#) = []
        pi(mark) = [1]
        pi(head) = []
        pi(a__head#) = []
        pi(nats) = []
        pi(a__nats#) = []
        pi(a__adx#) = []
        pi(a__zeros) = []
        pi(adx) = []
        pi(incr) = []
        pi(a__incr) = []
        pi(nil) = []
        pi(a__adx) = [1]
        pi(a__nats) = []
        pi(a__head) = []
        pi(a__tail) = [1]
        pi(|0|) = []
        pi(zeros) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__incr#_A(x1) = 45
          cons_A(x1,x2) = 54
          mark#_A(x1) = 45
          s_A(x1) = 45
          tail_A(x1) = 25
          a__tail#_A(x1) = 46
          mark_A(x1) = max{26, x1}
          head_A(x1) = 47
          a__head#_A(x1) = 44
          nats_A = 0
          a__nats#_A = 33
          a__adx#_A(x1) = 45
          a__zeros_A = 55
          adx_A(x1) = 25
          incr_A(x1) = 66
          a__incr_A(x1) = 66
          nil_A = 27
          a__adx_A(x1) = 26
          a__nats_A = 26
          a__head_A(x1) = 47
          a__tail_A(x1) = 26
          |0|_A = 56
          zeros_A = 55
    
      precedence:
      
        tail > nil > mark = a__nats# = a__zeros = a__nats = |0| = zeros > s = a__tail# > incr = a__incr = a__adx = a__head > cons > a__incr# = mark# = nats = a__adx# > a__head# = adx > a__tail > head
      
      partial status:
    
        pi(a__incr#) = []
        pi(cons) = []
        pi(mark#) = []
        pi(s) = []
        pi(tail) = []
        pi(a__tail#) = []
        pi(mark) = [1]
        pi(head) = []
        pi(a__head#) = []
        pi(nats) = []
        pi(a__nats#) = []
        pi(a__adx#) = []
        pi(a__zeros) = []
        pi(adx) = []
        pi(incr) = []
        pi(a__incr) = []
        pi(nil) = []
        pi(a__adx) = []
        pi(a__nats) = []
        pi(a__head) = []
        pi(a__tail) = []
        pi(|0|) = []
        pi(zeros) = []
    

The next rules are strictly ordered:

  p5, p9, p11, p12, p13

We remove them from the problem.

-- SCC decomposition.

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

p1: a__incr#(cons(X,L)) -> mark#(X)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(tail(X)) -> mark#(X)
p5: a__tail#(cons(X,L)) -> mark#(L)
p6: mark#(head(X)) -> mark#(X)
p7: mark#(head(X)) -> a__head#(mark(X))
p8: mark#(nats()) -> a__nats#()
p9: mark#(adx(X)) -> a__adx#(mark(X))
p10: a__adx#(cons(X,L)) -> a__incr#(cons(mark(X),adx(L)))
p11: mark#(incr(X)) -> mark#(X)
p12: mark#(incr(X)) -> a__incr#(mark(X))

and R consists of:

r1: a__incr(nil()) -> nil()
r2: a__incr(cons(X,L)) -> cons(s(mark(X)),incr(L))
r3: a__adx(nil()) -> nil()
r4: a__adx(cons(X,L)) -> a__incr(cons(mark(X),adx(L)))
r5: a__nats() -> a__adx(a__zeros())
r6: a__zeros() -> cons(|0|(),zeros())
r7: a__head(cons(X,L)) -> mark(X)
r8: a__tail(cons(X,L)) -> mark(L)
r9: mark(incr(X)) -> a__incr(mark(X))
r10: mark(adx(X)) -> a__adx(mark(X))
r11: mark(nats()) -> a__nats()
r12: mark(zeros()) -> a__zeros()
r13: mark(head(X)) -> a__head(mark(X))
r14: mark(tail(X)) -> a__tail(mark(X))
r15: mark(nil()) -> nil()
r16: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r17: mark(s(X)) -> s(mark(X))
r18: mark(|0|()) -> |0|()
r19: a__incr(X) -> incr(X)
r20: a__adx(X) -> adx(X)
r21: a__nats() -> nats()
r22: a__zeros() -> zeros()
r23: a__head(X) -> head(X)
r24: a__tail(X) -> tail(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

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

and R consists of:

r1: a__incr(nil()) -> nil()
r2: a__incr(cons(X,L)) -> cons(s(mark(X)),incr(L))
r3: a__adx(nil()) -> nil()
r4: a__adx(cons(X,L)) -> a__incr(cons(mark(X),adx(L)))
r5: a__nats() -> a__adx(a__zeros())
r6: a__zeros() -> cons(|0|(),zeros())
r7: a__head(cons(X,L)) -> mark(X)
r8: a__tail(cons(X,L)) -> mark(L)
r9: mark(incr(X)) -> a__incr(mark(X))
r10: mark(adx(X)) -> a__adx(mark(X))
r11: mark(nats()) -> a__nats()
r12: mark(zeros()) -> a__zeros()
r13: mark(head(X)) -> a__head(mark(X))
r14: mark(tail(X)) -> a__tail(mark(X))
r15: mark(nil()) -> nil()
r16: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r17: mark(s(X)) -> s(mark(X))
r18: mark(|0|()) -> |0|()
r19: a__incr(X) -> incr(X)
r20: a__adx(X) -> adx(X)
r21: a__nats() -> nats()
r22: a__zeros() -> zeros()
r23: a__head(X) -> head(X)
r24: a__tail(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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__incr#_A(x1) = max{2, x1}
          cons_A(x1,x2) = max{7, x1 + 5, x2}
          mark#_A(x1) = x1 + 4
          incr_A(x1) = max{1, x1}
          mark_A(x1) = x1
          adx_A(x1) = x1 + 10
          a__adx#_A(x1) = x1 + 10
          head_A(x1) = x1 + 5
          tail_A(x1) = x1 + 5
          s_A(x1) = max{2, x1}
          a__incr_A(x1) = max{1, x1}
          nil_A = 9
          a__adx_A(x1) = x1 + 10
          a__nats_A = 19
          a__zeros_A = 8
          |0|_A = 3
          zeros_A = 8
          a__head_A(x1) = x1 + 5
          a__tail_A(x1) = x1 + 5
          nats_A = 19
    
      precedence:
      
        adx = a__adx > mark# = a__adx# = a__zeros = zeros > mark = a__head = a__tail > a__nats > |0| = nats > tail > cons = s = a__incr > incr > a__incr# > head = nil
      
      partial status:
    
        pi(a__incr#) = [1]
        pi(cons) = [1]
        pi(mark#) = [1]
        pi(incr) = [1]
        pi(mark) = [1]
        pi(adx) = []
        pi(a__adx#) = [1]
        pi(head) = [1]
        pi(tail) = [1]
        pi(s) = [1]
        pi(a__incr) = [1]
        pi(nil) = []
        pi(a__adx) = []
        pi(a__nats) = []
        pi(a__zeros) = []
        pi(|0|) = []
        pi(zeros) = []
        pi(a__head) = [1]
        pi(a__tail) = []
        pi(nats) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__incr#_A(x1) = max{2, x1 - 14}
          cons_A(x1,x2) = max{22, x1 + 7}
          mark#_A(x1) = max{3, x1 - 8}
          incr_A(x1) = max{21, x1 - 7}
          mark_A(x1) = 27
          adx_A(x1) = 27
          a__adx#_A(x1) = max{1, x1}
          head_A(x1) = max{12, x1 + 11}
          tail_A(x1) = x1 + 12
          s_A(x1) = 0
          a__incr_A(x1) = max{23, x1 - 7}
          nil_A = 28
          a__adx_A(x1) = 27
          a__nats_A = 25
          a__zeros_A = 26
          |0|_A = 18
          zeros_A = 25
          a__head_A(x1) = x1 + 43
          a__tail_A(x1) = 28
          nats_A = 26
    
      precedence:
      
        a__incr = nil > tail = a__tail > mark > a__adx > adx > mark# = incr = a__adx# = s = a__nats = a__zeros = |0| = zeros = a__head = nats > a__incr# = cons = head
      
      partial status:
    
        pi(a__incr#) = []
        pi(cons) = [1]
        pi(mark#) = []
        pi(incr) = []
        pi(mark) = []
        pi(adx) = []
        pi(a__adx#) = []
        pi(head) = [1]
        pi(tail) = [1]
        pi(s) = []
        pi(a__incr) = []
        pi(nil) = []
        pi(a__adx) = []
        pi(a__nats) = []
        pi(a__zeros) = []
        pi(|0|) = []
        pi(zeros) = []
        pi(a__head) = []
        pi(a__tail) = []
        pi(nats) = []
    

The next rules are strictly ordered:

  p3, p4, p6, p7, p8, p9

We remove them from the problem.

-- SCC decomposition.

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

p1: a__incr#(cons(X,L)) -> mark#(X)
p2: mark#(incr(X)) -> a__incr#(mark(X))
p3: a__adx#(cons(X,L)) -> a__incr#(cons(mark(X),adx(L)))

and R consists of:

r1: a__incr(nil()) -> nil()
r2: a__incr(cons(X,L)) -> cons(s(mark(X)),incr(L))
r3: a__adx(nil()) -> nil()
r4: a__adx(cons(X,L)) -> a__incr(cons(mark(X),adx(L)))
r5: a__nats() -> a__adx(a__zeros())
r6: a__zeros() -> cons(|0|(),zeros())
r7: a__head(cons(X,L)) -> mark(X)
r8: a__tail(cons(X,L)) -> mark(L)
r9: mark(incr(X)) -> a__incr(mark(X))
r10: mark(adx(X)) -> a__adx(mark(X))
r11: mark(nats()) -> a__nats()
r12: mark(zeros()) -> a__zeros()
r13: mark(head(X)) -> a__head(mark(X))
r14: mark(tail(X)) -> a__tail(mark(X))
r15: mark(nil()) -> nil()
r16: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r17: mark(s(X)) -> s(mark(X))
r18: mark(|0|()) -> |0|()
r19: a__incr(X) -> incr(X)
r20: a__adx(X) -> adx(X)
r21: a__nats() -> nats()
r22: a__zeros() -> zeros()
r23: a__head(X) -> head(X)
r24: a__tail(X) -> tail(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: a__incr#(cons(X,L)) -> mark#(X)
p2: mark#(incr(X)) -> a__incr#(mark(X))

and R consists of:

r1: a__incr(nil()) -> nil()
r2: a__incr(cons(X,L)) -> cons(s(mark(X)),incr(L))
r3: a__adx(nil()) -> nil()
r4: a__adx(cons(X,L)) -> a__incr(cons(mark(X),adx(L)))
r5: a__nats() -> a__adx(a__zeros())
r6: a__zeros() -> cons(|0|(),zeros())
r7: a__head(cons(X,L)) -> mark(X)
r8: a__tail(cons(X,L)) -> mark(L)
r9: mark(incr(X)) -> a__incr(mark(X))
r10: mark(adx(X)) -> a__adx(mark(X))
r11: mark(nats()) -> a__nats()
r12: mark(zeros()) -> a__zeros()
r13: mark(head(X)) -> a__head(mark(X))
r14: mark(tail(X)) -> a__tail(mark(X))
r15: mark(nil()) -> nil()
r16: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r17: mark(s(X)) -> s(mark(X))
r18: mark(|0|()) -> |0|()
r19: a__incr(X) -> incr(X)
r20: a__adx(X) -> adx(X)
r21: a__nats() -> nats()
r22: a__zeros() -> zeros()
r23: a__head(X) -> head(X)
r24: a__tail(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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__incr#_A(x1) = max{8, x1 - 11}
          cons_A(x1,x2) = max{x1 + 23, x2 + 8}
          mark#_A(x1) = x1 + 11
          incr_A(x1) = max{22, x1}
          mark_A(x1) = x1 + 10
          a__incr_A(x1) = max{30, x1}
          nil_A = 0
          s_A(x1) = 0
          a__adx_A(x1) = x1 + 30
          adx_A(x1) = x1 + 30
          a__nats_A = 55
          a__zeros_A = 24
          |0|_A = 0
          zeros_A = 15
          a__head_A(x1) = max{10, x1 - 11}
          a__tail_A(x1) = max{14, x1 + 3}
          nats_A = 46
          head_A(x1) = max{0, x1 - 11}
          tail_A(x1) = max{4, x1 + 3}
    
      precedence:
      
        nil > mark = a__zeros > a__nats > a__adx > a__incr > incr > a__incr# > zeros > adx > mark# = a__tail > tail > cons = nats > s = |0| = a__head = head
      
      partial status:
    
        pi(a__incr#) = []
        pi(cons) = [1, 2]
        pi(mark#) = [1]
        pi(incr) = [1]
        pi(mark) = [1]
        pi(a__incr) = [1]
        pi(nil) = []
        pi(s) = []
        pi(a__adx) = []
        pi(adx) = []
        pi(a__nats) = []
        pi(a__zeros) = []
        pi(|0|) = []
        pi(zeros) = []
        pi(a__head) = []
        pi(a__tail) = [1]
        pi(nats) = []
        pi(head) = []
        pi(tail) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__incr#_A(x1) = 7
          cons_A(x1,x2) = max{x1 + 28, x2 + 40}
          mark#_A(x1) = max{3, x1 + 2}
          incr_A(x1) = max{8, x1 + 6}
          mark_A(x1) = max{5, x1}
          a__incr_A(x1) = x1 + 21
          nil_A = 6
          s_A(x1) = 11
          a__adx_A(x1) = 27
          adx_A(x1) = 27
          a__nats_A = 28
          a__zeros_A = 3
          |0|_A = 6
          zeros_A = 4
          a__head_A(x1) = 4
          a__tail_A(x1) = x1 + 7
          nats_A = 29
          head_A(x1) = 4
          tail_A(x1) = x1 + 13
    
      precedence:
      
        s = a__nats = a__zeros = |0| = zeros = nats > a__adx > a__incr > a__incr# = cons = mark# = mark = nil = a__head = head > incr = adx = a__tail = tail
      
      partial status:
    
        pi(a__incr#) = []
        pi(cons) = [2]
        pi(mark#) = [1]
        pi(incr) = []
        pi(mark) = [1]
        pi(a__incr) = [1]
        pi(nil) = []
        pi(s) = []
        pi(a__adx) = []
        pi(adx) = []
        pi(a__nats) = []
        pi(a__zeros) = []
        pi(|0|) = []
        pi(zeros) = []
        pi(a__head) = []
        pi(a__tail) = []
        pi(nats) = []
        pi(head) = []
        pi(tail) = []
    

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#(incr(X)) -> a__incr#(mark(X))

and R consists of:

r1: a__incr(nil()) -> nil()
r2: a__incr(cons(X,L)) -> cons(s(mark(X)),incr(L))
r3: a__adx(nil()) -> nil()
r4: a__adx(cons(X,L)) -> a__incr(cons(mark(X),adx(L)))
r5: a__nats() -> a__adx(a__zeros())
r6: a__zeros() -> cons(|0|(),zeros())
r7: a__head(cons(X,L)) -> mark(X)
r8: a__tail(cons(X,L)) -> mark(L)
r9: mark(incr(X)) -> a__incr(mark(X))
r10: mark(adx(X)) -> a__adx(mark(X))
r11: mark(nats()) -> a__nats()
r12: mark(zeros()) -> a__zeros()
r13: mark(head(X)) -> a__head(mark(X))
r14: mark(tail(X)) -> a__tail(mark(X))
r15: mark(nil()) -> nil()
r16: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r17: mark(s(X)) -> s(mark(X))
r18: mark(|0|()) -> |0|()
r19: a__incr(X) -> incr(X)
r20: a__adx(X) -> adx(X)
r21: a__nats() -> nats()
r22: a__zeros() -> zeros()
r23: a__head(X) -> head(X)
r24: a__tail(X) -> tail(X)

The estimated dependency graph contains the following SCCs:

  (no SCCs)