YES

We show the termination of the TRS R:

  a__nats() -> a__adx(a__zeros())
  a__zeros() -> cons(|0|(),zeros())
  a__incr(cons(X,Y)) -> cons(s(X),incr(Y))
  a__adx(cons(X,Y)) -> a__incr(cons(X,adx(Y)))
  a__hd(cons(X,Y)) -> mark(X)
  a__tl(cons(X,Y)) -> mark(Y)
  mark(nats()) -> a__nats()
  mark(adx(X)) -> a__adx(mark(X))
  mark(zeros()) -> a__zeros()
  mark(incr(X)) -> a__incr(mark(X))
  mark(hd(X)) -> a__hd(mark(X))
  mark(tl(X)) -> a__tl(mark(X))
  mark(cons(X1,X2)) -> cons(X1,X2)
  mark(|0|()) -> |0|()
  mark(s(X)) -> s(X)
  a__nats() -> nats()
  a__adx(X) -> adx(X)
  a__zeros() -> zeros()
  a__incr(X) -> incr(X)
  a__hd(X) -> hd(X)
  a__tl(X) -> tl(X)

-- SCC decomposition.

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

p1: a__nats#() -> a__adx#(a__zeros())
p2: a__nats#() -> a__zeros#()
p3: a__adx#(cons(X,Y)) -> a__incr#(cons(X,adx(Y)))
p4: a__hd#(cons(X,Y)) -> mark#(X)
p5: a__tl#(cons(X,Y)) -> mark#(Y)
p6: mark#(nats()) -> a__nats#()
p7: mark#(adx(X)) -> a__adx#(mark(X))
p8: mark#(adx(X)) -> mark#(X)
p9: mark#(zeros()) -> a__zeros#()
p10: mark#(incr(X)) -> a__incr#(mark(X))
p11: mark#(incr(X)) -> mark#(X)
p12: mark#(hd(X)) -> a__hd#(mark(X))
p13: mark#(hd(X)) -> mark#(X)
p14: mark#(tl(X)) -> a__tl#(mark(X))
p15: mark#(tl(X)) -> mark#(X)

and R consists of:

r1: a__nats() -> a__adx(a__zeros())
r2: a__zeros() -> cons(|0|(),zeros())
r3: a__incr(cons(X,Y)) -> cons(s(X),incr(Y))
r4: a__adx(cons(X,Y)) -> a__incr(cons(X,adx(Y)))
r5: a__hd(cons(X,Y)) -> mark(X)
r6: a__tl(cons(X,Y)) -> mark(Y)
r7: mark(nats()) -> a__nats()
r8: mark(adx(X)) -> a__adx(mark(X))
r9: mark(zeros()) -> a__zeros()
r10: mark(incr(X)) -> a__incr(mark(X))
r11: mark(hd(X)) -> a__hd(mark(X))
r12: mark(tl(X)) -> a__tl(mark(X))
r13: mark(cons(X1,X2)) -> cons(X1,X2)
r14: mark(|0|()) -> |0|()
r15: mark(s(X)) -> s(X)
r16: a__nats() -> nats()
r17: a__adx(X) -> adx(X)
r18: a__zeros() -> zeros()
r19: a__incr(X) -> incr(X)
r20: a__hd(X) -> hd(X)
r21: a__tl(X) -> tl(X)

The estimated dependency graph contains the following SCCs:

  {p4, p5, p8, p11, p12, p13, p14, p15}


-- Reduction pair.

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

p1: a__tl#(cons(X,Y)) -> mark#(Y)
p2: mark#(tl(X)) -> mark#(X)
p3: mark#(tl(X)) -> a__tl#(mark(X))
p4: mark#(hd(X)) -> mark#(X)
p5: mark#(hd(X)) -> a__hd#(mark(X))
p6: a__hd#(cons(X,Y)) -> mark#(X)
p7: mark#(incr(X)) -> mark#(X)
p8: mark#(adx(X)) -> mark#(X)

and R consists of:

r1: a__nats() -> a__adx(a__zeros())
r2: a__zeros() -> cons(|0|(),zeros())
r3: a__incr(cons(X,Y)) -> cons(s(X),incr(Y))
r4: a__adx(cons(X,Y)) -> a__incr(cons(X,adx(Y)))
r5: a__hd(cons(X,Y)) -> mark(X)
r6: a__tl(cons(X,Y)) -> mark(Y)
r7: mark(nats()) -> a__nats()
r8: mark(adx(X)) -> a__adx(mark(X))
r9: mark(zeros()) -> a__zeros()
r10: mark(incr(X)) -> a__incr(mark(X))
r11: mark(hd(X)) -> a__hd(mark(X))
r12: mark(tl(X)) -> a__tl(mark(X))
r13: mark(cons(X1,X2)) -> cons(X1,X2)
r14: mark(|0|()) -> |0|()
r15: mark(s(X)) -> s(X)
r16: a__nats() -> nats()
r17: a__adx(X) -> adx(X)
r18: a__zeros() -> zeros()
r19: a__incr(X) -> incr(X)
r20: a__hd(X) -> hd(X)
r21: a__tl(X) -> tl(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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          a__tl#_A(x1) = x1
          cons_A(x1,x2) = x1 + x2 + 2
          mark#_A(x1) = x1
          tl_A(x1) = x1 + 6
          mark_A(x1) = x1 + 5
          hd_A(x1) = x1 + 7
          a__hd#_A(x1) = x1 + 1
          incr_A(x1) = x1
          adx_A(x1) = x1 + 1
          a__nats_A() = 7
          a__adx_A(x1) = x1 + 1
          a__zeros_A() = 5
          |0|_A() = 1
          zeros_A() = 1
          a__incr_A(x1) = x1
          s_A(x1) = x1
          a__hd_A(x1) = x1 + 7
          a__tl_A(x1) = x1 + 6
          nats_A() = 3
  
    precedence:
    
      a__zeros > |0| = zeros > mark > adx = a__adx = a__hd > cons = incr = a__incr > a__tl# = mark# = a__hd# > a__nats = nats > hd > tl = s = a__tl
    
    partial status:
  
      pi(a__tl#) = [1]
      pi(cons) = [1, 2]
      pi(mark#) = [1]
      pi(tl) = []
      pi(mark) = [1]
      pi(hd) = []
      pi(a__hd#) = [1]
      pi(incr) = [1]
      pi(adx) = []
      pi(a__nats) = []
      pi(a__adx) = []
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__incr) = [1]
      pi(s) = [1]
      pi(a__hd) = []
      pi(a__tl) = [1]
      pi(nats) = []

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: a__tl#(cons(X,Y)) -> mark#(Y)
p2: mark#(tl(X)) -> mark#(X)
p3: mark#(tl(X)) -> a__tl#(mark(X))
p4: mark#(hd(X)) -> mark#(X)
p5: mark#(hd(X)) -> a__hd#(mark(X))
p6: mark#(incr(X)) -> mark#(X)
p7: mark#(adx(X)) -> mark#(X)

and R consists of:

r1: a__nats() -> a__adx(a__zeros())
r2: a__zeros() -> cons(|0|(),zeros())
r3: a__incr(cons(X,Y)) -> cons(s(X),incr(Y))
r4: a__adx(cons(X,Y)) -> a__incr(cons(X,adx(Y)))
r5: a__hd(cons(X,Y)) -> mark(X)
r6: a__tl(cons(X,Y)) -> mark(Y)
r7: mark(nats()) -> a__nats()
r8: mark(adx(X)) -> a__adx(mark(X))
r9: mark(zeros()) -> a__zeros()
r10: mark(incr(X)) -> a__incr(mark(X))
r11: mark(hd(X)) -> a__hd(mark(X))
r12: mark(tl(X)) -> a__tl(mark(X))
r13: mark(cons(X1,X2)) -> cons(X1,X2)
r14: mark(|0|()) -> |0|()
r15: mark(s(X)) -> s(X)
r16: a__nats() -> nats()
r17: a__adx(X) -> adx(X)
r18: a__zeros() -> zeros()
r19: a__incr(X) -> incr(X)
r20: a__hd(X) -> hd(X)
r21: a__tl(X) -> tl(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p6, p7}


-- Reduction pair.

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

p1: a__tl#(cons(X,Y)) -> mark#(Y)
p2: mark#(adx(X)) -> mark#(X)
p3: mark#(incr(X)) -> mark#(X)
p4: mark#(hd(X)) -> mark#(X)
p5: mark#(tl(X)) -> a__tl#(mark(X))
p6: mark#(tl(X)) -> mark#(X)

and R consists of:

r1: a__nats() -> a__adx(a__zeros())
r2: a__zeros() -> cons(|0|(),zeros())
r3: a__incr(cons(X,Y)) -> cons(s(X),incr(Y))
r4: a__adx(cons(X,Y)) -> a__incr(cons(X,adx(Y)))
r5: a__hd(cons(X,Y)) -> mark(X)
r6: a__tl(cons(X,Y)) -> mark(Y)
r7: mark(nats()) -> a__nats()
r8: mark(adx(X)) -> a__adx(mark(X))
r9: mark(zeros()) -> a__zeros()
r10: mark(incr(X)) -> a__incr(mark(X))
r11: mark(hd(X)) -> a__hd(mark(X))
r12: mark(tl(X)) -> a__tl(mark(X))
r13: mark(cons(X1,X2)) -> cons(X1,X2)
r14: mark(|0|()) -> |0|()
r15: mark(s(X)) -> s(X)
r16: a__nats() -> nats()
r17: a__adx(X) -> adx(X)
r18: a__zeros() -> zeros()
r19: a__incr(X) -> incr(X)
r20: a__hd(X) -> hd(X)
r21: a__tl(X) -> tl(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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          a__tl#_A(x1) = x1 + 1
          cons_A(x1,x2) = x1 + x2 + 1
          mark#_A(x1) = x1
          adx_A(x1) = x1
          incr_A(x1) = x1
          hd_A(x1) = x1 + 5
          tl_A(x1) = x1 + 6
          mark_A(x1) = x1 + 4
          a__nats_A() = 5
          a__adx_A(x1) = x1
          a__zeros_A() = 4
          |0|_A() = 1
          zeros_A() = 1
          a__incr_A(x1) = x1
          s_A(x1) = x1
          a__hd_A(x1) = x1 + 5
          a__tl_A(x1) = x1 + 6
          nats_A() = 2
  
    precedence:
    
      cons = zeros > a__tl# = mark# > |0| = s > mark > a__hd > adx = a__adx > hd = tl = a__tl > a__nats > a__zeros = nats > a__incr > incr
    
    partial status:
  
      pi(a__tl#) = [1]
      pi(cons) = []
      pi(mark#) = [1]
      pi(adx) = [1]
      pi(incr) = [1]
      pi(hd) = [1]
      pi(tl) = []
      pi(mark) = [1]
      pi(a__nats) = []
      pi(a__adx) = [1]
      pi(a__zeros) = []
      pi(|0|) = []
      pi(zeros) = []
      pi(a__incr) = [1]
      pi(s) = []
      pi(a__hd) = []
      pi(a__tl) = [1]
      pi(nats) = []

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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

p1: a__tl#(cons(X,Y)) -> mark#(Y)
p2: mark#(adx(X)) -> mark#(X)
p3: mark#(incr(X)) -> mark#(X)
p4: mark#(hd(X)) -> mark#(X)
p5: mark#(tl(X)) -> mark#(X)

and R consists of:

r1: a__nats() -> a__adx(a__zeros())
r2: a__zeros() -> cons(|0|(),zeros())
r3: a__incr(cons(X,Y)) -> cons(s(X),incr(Y))
r4: a__adx(cons(X,Y)) -> a__incr(cons(X,adx(Y)))
r5: a__hd(cons(X,Y)) -> mark(X)
r6: a__tl(cons(X,Y)) -> mark(Y)
r7: mark(nats()) -> a__nats()
r8: mark(adx(X)) -> a__adx(mark(X))
r9: mark(zeros()) -> a__zeros()
r10: mark(incr(X)) -> a__incr(mark(X))
r11: mark(hd(X)) -> a__hd(mark(X))
r12: mark(tl(X)) -> a__tl(mark(X))
r13: mark(cons(X1,X2)) -> cons(X1,X2)
r14: mark(|0|()) -> |0|()
r15: mark(s(X)) -> s(X)
r16: a__nats() -> nats()
r17: a__adx(X) -> adx(X)
r18: a__zeros() -> zeros()
r19: a__incr(X) -> incr(X)
r20: a__hd(X) -> hd(X)
r21: a__tl(X) -> tl(X)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p5}


-- Reduction pair.

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

p1: mark#(adx(X)) -> mark#(X)
p2: mark#(tl(X)) -> mark#(X)
p3: mark#(hd(X)) -> mark#(X)
p4: mark#(incr(X)) -> mark#(X)

and R consists of:

r1: a__nats() -> a__adx(a__zeros())
r2: a__zeros() -> cons(|0|(),zeros())
r3: a__incr(cons(X,Y)) -> cons(s(X),incr(Y))
r4: a__adx(cons(X,Y)) -> a__incr(cons(X,adx(Y)))
r5: a__hd(cons(X,Y)) -> mark(X)
r6: a__tl(cons(X,Y)) -> mark(Y)
r7: mark(nats()) -> a__nats()
r8: mark(adx(X)) -> a__adx(mark(X))
r9: mark(zeros()) -> a__zeros()
r10: mark(incr(X)) -> a__incr(mark(X))
r11: mark(hd(X)) -> a__hd(mark(X))
r12: mark(tl(X)) -> a__tl(mark(X))
r13: mark(cons(X1,X2)) -> cons(X1,X2)
r14: mark(|0|()) -> |0|()
r15: mark(s(X)) -> s(X)
r16: a__nats() -> nats()
r17: a__adx(X) -> adx(X)
r18: a__zeros() -> zeros()
r19: a__incr(X) -> incr(X)
r20: a__hd(X) -> hd(X)
r21: a__tl(X) -> tl(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1
          adx_A(x1) = x1 + 1
          tl_A(x1) = x1 + 1
          hd_A(x1) = x1 + 1
          incr_A(x1) = x1 + 1
  
    precedence:
    
      mark# = adx = tl = incr > hd
    
    partial status:
  
      pi(mark#) = [1]
      pi(adx) = [1]
      pi(tl) = [1]
      pi(hd) = []
      pi(incr) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(tl(X)) -> mark#(X)
p2: mark#(hd(X)) -> mark#(X)
p3: mark#(incr(X)) -> mark#(X)

and R consists of:

r1: a__nats() -> a__adx(a__zeros())
r2: a__zeros() -> cons(|0|(),zeros())
r3: a__incr(cons(X,Y)) -> cons(s(X),incr(Y))
r4: a__adx(cons(X,Y)) -> a__incr(cons(X,adx(Y)))
r5: a__hd(cons(X,Y)) -> mark(X)
r6: a__tl(cons(X,Y)) -> mark(Y)
r7: mark(nats()) -> a__nats()
r8: mark(adx(X)) -> a__adx(mark(X))
r9: mark(zeros()) -> a__zeros()
r10: mark(incr(X)) -> a__incr(mark(X))
r11: mark(hd(X)) -> a__hd(mark(X))
r12: mark(tl(X)) -> a__tl(mark(X))
r13: mark(cons(X1,X2)) -> cons(X1,X2)
r14: mark(|0|()) -> |0|()
r15: mark(s(X)) -> s(X)
r16: a__nats() -> nats()
r17: a__adx(X) -> adx(X)
r18: a__zeros() -> zeros()
r19: a__incr(X) -> incr(X)
r20: a__hd(X) -> hd(X)
r21: a__tl(X) -> tl(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#(tl(X)) -> mark#(X)
p2: mark#(incr(X)) -> mark#(X)
p3: mark#(hd(X)) -> mark#(X)

and R consists of:

r1: a__nats() -> a__adx(a__zeros())
r2: a__zeros() -> cons(|0|(),zeros())
r3: a__incr(cons(X,Y)) -> cons(s(X),incr(Y))
r4: a__adx(cons(X,Y)) -> a__incr(cons(X,adx(Y)))
r5: a__hd(cons(X,Y)) -> mark(X)
r6: a__tl(cons(X,Y)) -> mark(Y)
r7: mark(nats()) -> a__nats()
r8: mark(adx(X)) -> a__adx(mark(X))
r9: mark(zeros()) -> a__zeros()
r10: mark(incr(X)) -> a__incr(mark(X))
r11: mark(hd(X)) -> a__hd(mark(X))
r12: mark(tl(X)) -> a__tl(mark(X))
r13: mark(cons(X1,X2)) -> cons(X1,X2)
r14: mark(|0|()) -> |0|()
r15: mark(s(X)) -> s(X)
r16: a__nats() -> nats()
r17: a__adx(X) -> adx(X)
r18: a__zeros() -> zeros()
r19: a__incr(X) -> incr(X)
r20: a__hd(X) -> hd(X)
r21: a__tl(X) -> tl(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1
          tl_A(x1) = x1 + 1
          incr_A(x1) = x1 + 1
          hd_A(x1) = x1 + 1
  
    precedence:
    
      mark# = tl = incr = hd
    
    partial status:
  
      pi(mark#) = [1]
      pi(tl) = [1]
      pi(incr) = [1]
      pi(hd) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(incr(X)) -> mark#(X)
p2: mark#(hd(X)) -> mark#(X)

and R consists of:

r1: a__nats() -> a__adx(a__zeros())
r2: a__zeros() -> cons(|0|(),zeros())
r3: a__incr(cons(X,Y)) -> cons(s(X),incr(Y))
r4: a__adx(cons(X,Y)) -> a__incr(cons(X,adx(Y)))
r5: a__hd(cons(X,Y)) -> mark(X)
r6: a__tl(cons(X,Y)) -> mark(Y)
r7: mark(nats()) -> a__nats()
r8: mark(adx(X)) -> a__adx(mark(X))
r9: mark(zeros()) -> a__zeros()
r10: mark(incr(X)) -> a__incr(mark(X))
r11: mark(hd(X)) -> a__hd(mark(X))
r12: mark(tl(X)) -> a__tl(mark(X))
r13: mark(cons(X1,X2)) -> cons(X1,X2)
r14: mark(|0|()) -> |0|()
r15: mark(s(X)) -> s(X)
r16: a__nats() -> nats()
r17: a__adx(X) -> adx(X)
r18: a__zeros() -> zeros()
r19: a__incr(X) -> incr(X)
r20: a__hd(X) -> hd(X)
r21: a__tl(X) -> tl(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#(incr(X)) -> mark#(X)
p2: mark#(hd(X)) -> mark#(X)

and R consists of:

r1: a__nats() -> a__adx(a__zeros())
r2: a__zeros() -> cons(|0|(),zeros())
r3: a__incr(cons(X,Y)) -> cons(s(X),incr(Y))
r4: a__adx(cons(X,Y)) -> a__incr(cons(X,adx(Y)))
r5: a__hd(cons(X,Y)) -> mark(X)
r6: a__tl(cons(X,Y)) -> mark(Y)
r7: mark(nats()) -> a__nats()
r8: mark(adx(X)) -> a__adx(mark(X))
r9: mark(zeros()) -> a__zeros()
r10: mark(incr(X)) -> a__incr(mark(X))
r11: mark(hd(X)) -> a__hd(mark(X))
r12: mark(tl(X)) -> a__tl(mark(X))
r13: mark(cons(X1,X2)) -> cons(X1,X2)
r14: mark(|0|()) -> |0|()
r15: mark(s(X)) -> s(X)
r16: a__nats() -> nats()
r17: a__adx(X) -> adx(X)
r18: a__zeros() -> zeros()
r19: a__incr(X) -> incr(X)
r20: a__hd(X) -> hd(X)
r21: a__tl(X) -> tl(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1
          incr_A(x1) = x1 + 1
          hd_A(x1) = x1 + 1
  
    precedence:
    
      mark# = incr = hd
    
    partial status:
  
      pi(mark#) = [1]
      pi(incr) = [1]
      pi(hd) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: a__nats() -> a__adx(a__zeros())
r2: a__zeros() -> cons(|0|(),zeros())
r3: a__incr(cons(X,Y)) -> cons(s(X),incr(Y))
r4: a__adx(cons(X,Y)) -> a__incr(cons(X,adx(Y)))
r5: a__hd(cons(X,Y)) -> mark(X)
r6: a__tl(cons(X,Y)) -> mark(Y)
r7: mark(nats()) -> a__nats()
r8: mark(adx(X)) -> a__adx(mark(X))
r9: mark(zeros()) -> a__zeros()
r10: mark(incr(X)) -> a__incr(mark(X))
r11: mark(hd(X)) -> a__hd(mark(X))
r12: mark(tl(X)) -> a__tl(mark(X))
r13: mark(cons(X1,X2)) -> cons(X1,X2)
r14: mark(|0|()) -> |0|()
r15: mark(s(X)) -> s(X)
r16: a__nats() -> nats()
r17: a__adx(X) -> adx(X)
r18: a__zeros() -> zeros()
r19: a__incr(X) -> incr(X)
r20: a__hd(X) -> hd(X)
r21: a__tl(X) -> tl(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#(hd(X)) -> mark#(X)

and R consists of:

r1: a__nats() -> a__adx(a__zeros())
r2: a__zeros() -> cons(|0|(),zeros())
r3: a__incr(cons(X,Y)) -> cons(s(X),incr(Y))
r4: a__adx(cons(X,Y)) -> a__incr(cons(X,adx(Y)))
r5: a__hd(cons(X,Y)) -> mark(X)
r6: a__tl(cons(X,Y)) -> mark(Y)
r7: mark(nats()) -> a__nats()
r8: mark(adx(X)) -> a__adx(mark(X))
r9: mark(zeros()) -> a__zeros()
r10: mark(incr(X)) -> a__incr(mark(X))
r11: mark(hd(X)) -> a__hd(mark(X))
r12: mark(tl(X)) -> a__tl(mark(X))
r13: mark(cons(X1,X2)) -> cons(X1,X2)
r14: mark(|0|()) -> |0|()
r15: mark(s(X)) -> s(X)
r16: a__nats() -> nats()
r17: a__adx(X) -> adx(X)
r18: a__zeros() -> zeros()
r19: a__incr(X) -> incr(X)
r20: a__hd(X) -> hd(X)
r21: a__tl(X) -> tl(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1
          hd_A(x1) = x1 + 1
  
    precedence:
    
      mark# = hd
    
    partial status:
  
      pi(mark#) = [1]
      pi(hd) = [1]

The next rules are strictly ordered:

  p1

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