YES

We show the termination of the TRS R:

  a__nats() -> cons(|0|(),incr(nats()))
  a__pairs() -> cons(|0|(),incr(odds()))
  a__odds() -> a__incr(a__pairs())
  a__incr(cons(X,XS)) -> cons(s(mark(X)),incr(XS))
  a__head(cons(X,XS)) -> mark(X)
  a__tail(cons(X,XS)) -> mark(XS)
  mark(nats()) -> a__nats()
  mark(incr(X)) -> a__incr(mark(X))
  mark(pairs()) -> a__pairs()
  mark(odds()) -> a__odds()
  mark(head(X)) -> a__head(mark(X))
  mark(tail(X)) -> a__tail(mark(X))
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(|0|()) -> |0|()
  mark(s(X)) -> s(mark(X))
  a__nats() -> nats()
  a__incr(X) -> incr(X)
  a__pairs() -> pairs()
  a__odds() -> odds()
  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__odds#() -> a__incr#(a__pairs())
p2: a__odds#() -> a__pairs#()
p3: a__incr#(cons(X,XS)) -> mark#(X)
p4: a__head#(cons(X,XS)) -> mark#(X)
p5: a__tail#(cons(X,XS)) -> mark#(XS)
p6: mark#(nats()) -> a__nats#()
p7: mark#(incr(X)) -> a__incr#(mark(X))
p8: mark#(incr(X)) -> mark#(X)
p9: mark#(pairs()) -> a__pairs#()
p10: mark#(odds()) -> a__odds#()
p11: mark#(head(X)) -> a__head#(mark(X))
p12: mark#(head(X)) -> mark#(X)
p13: mark#(tail(X)) -> a__tail#(mark(X))
p14: mark#(tail(X)) -> mark#(X)
p15: mark#(cons(X1,X2)) -> mark#(X1)
p16: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__nats() -> cons(|0|(),incr(nats()))
r2: a__pairs() -> cons(|0|(),incr(odds()))
r3: a__odds() -> a__incr(a__pairs())
r4: a__incr(cons(X,XS)) -> cons(s(mark(X)),incr(XS))
r5: a__head(cons(X,XS)) -> mark(X)
r6: a__tail(cons(X,XS)) -> mark(XS)
r7: mark(nats()) -> a__nats()
r8: mark(incr(X)) -> a__incr(mark(X))
r9: mark(pairs()) -> a__pairs()
r10: mark(odds()) -> a__odds()
r11: mark(head(X)) -> a__head(mark(X))
r12: mark(tail(X)) -> a__tail(mark(X))
r13: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r14: mark(|0|()) -> |0|()
r15: mark(s(X)) -> s(mark(X))
r16: a__nats() -> nats()
r17: a__incr(X) -> incr(X)
r18: a__pairs() -> pairs()
r19: a__odds() -> odds()
r20: a__head(X) -> head(X)
r21: a__tail(X) -> tail(X)

The estimated dependency graph contains the following SCCs:

  {p1, p3, p4, p5, p7, p8, p10, p11, p12, p13, p14, p15, p16}


-- Reduction pair.

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

p1: a__odds#() -> a__incr#(a__pairs())
p2: a__incr#(cons(X,XS)) -> mark#(X)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: mark#(tail(X)) -> mark#(X)
p6: mark#(tail(X)) -> a__tail#(mark(X))
p7: a__tail#(cons(X,XS)) -> mark#(XS)
p8: mark#(head(X)) -> mark#(X)
p9: mark#(head(X)) -> a__head#(mark(X))
p10: a__head#(cons(X,XS)) -> mark#(X)
p11: mark#(odds()) -> a__odds#()
p12: mark#(incr(X)) -> mark#(X)
p13: mark#(incr(X)) -> a__incr#(mark(X))

and R consists of:

r1: a__nats() -> cons(|0|(),incr(nats()))
r2: a__pairs() -> cons(|0|(),incr(odds()))
r3: a__odds() -> a__incr(a__pairs())
r4: a__incr(cons(X,XS)) -> cons(s(mark(X)),incr(XS))
r5: a__head(cons(X,XS)) -> mark(X)
r6: a__tail(cons(X,XS)) -> mark(XS)
r7: mark(nats()) -> a__nats()
r8: mark(incr(X)) -> a__incr(mark(X))
r9: mark(pairs()) -> a__pairs()
r10: mark(odds()) -> a__odds()
r11: mark(head(X)) -> a__head(mark(X))
r12: mark(tail(X)) -> a__tail(mark(X))
r13: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r14: mark(|0|()) -> |0|()
r15: mark(s(X)) -> s(mark(X))
r16: a__nats() -> nats()
r17: a__incr(X) -> incr(X)
r18: a__pairs() -> pairs()
r19: a__odds() -> odds()
r20: a__head(X) -> head(X)
r21: 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

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      a__odds#_A() = 1
      a__incr#_A(x1) = x1
      a__pairs_A() = 1
      cons_A(x1,x2) = x1 + x2
      mark#_A(x1) = x1
      s_A(x1) = x1
      tail_A(x1) = x1 + 1
      a__tail#_A(x1) = x1 + 1
      mark_A(x1) = x1
      head_A(x1) = x1 + 2
      a__head#_A(x1) = x1 + 1
      odds_A() = 1
      incr_A(x1) = x1
      a__nats_A() = 1
      |0|_A() = 0
      nats_A() = 1
      a__odds_A() = 1
      a__incr_A(x1) = x1
      a__head_A(x1) = x1 + 2
      a__tail_A(x1) = x1 + 1
      pairs_A() = 1

The next rules are strictly ordered:

  p5, p7, p8, p9, p10
  r5, r6

We remove them from the problem.

-- SCC decomposition.

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

p1: a__odds#() -> a__incr#(a__pairs())
p2: a__incr#(cons(X,XS)) -> mark#(X)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: mark#(tail(X)) -> a__tail#(mark(X))
p6: mark#(odds()) -> a__odds#()
p7: mark#(incr(X)) -> mark#(X)
p8: mark#(incr(X)) -> a__incr#(mark(X))

and R consists of:

r1: a__nats() -> cons(|0|(),incr(nats()))
r2: a__odds() -> a__incr(a__pairs())
r3: a__incr(cons(X,XS)) -> cons(s(mark(X)),incr(XS))
r4: a__nats() -> nats()
r5: a__incr(X) -> incr(X)
r6: a__odds() -> odds()
r7: a__head(X) -> head(X)
r8: a__tail(X) -> tail(X)
r9: a__pairs() -> cons(|0|(),incr(odds()))
r10: mark(nats()) -> a__nats()
r11: mark(incr(X)) -> a__incr(mark(X))
r12: mark(pairs()) -> a__pairs()
r13: mark(odds()) -> a__odds()
r14: mark(head(X)) -> a__head(mark(X))
r15: mark(tail(X)) -> a__tail(mark(X))
r16: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r17: mark(|0|()) -> |0|()
r18: mark(s(X)) -> s(mark(X))
r19: a__pairs() -> pairs()

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: a__odds#() -> a__incr#(a__pairs())
p2: a__incr#(cons(X,XS)) -> mark#(X)
p3: mark#(incr(X)) -> a__incr#(mark(X))
p4: mark#(incr(X)) -> mark#(X)
p5: mark#(odds()) -> a__odds#()
p6: mark#(cons(X1,X2)) -> mark#(X1)
p7: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__nats() -> cons(|0|(),incr(nats()))
r2: a__odds() -> a__incr(a__pairs())
r3: a__incr(cons(X,XS)) -> cons(s(mark(X)),incr(XS))
r4: a__nats() -> nats()
r5: a__incr(X) -> incr(X)
r6: a__odds() -> odds()
r7: a__head(X) -> head(X)
r8: a__tail(X) -> tail(X)
r9: a__pairs() -> cons(|0|(),incr(odds()))
r10: mark(nats()) -> a__nats()
r11: mark(incr(X)) -> a__incr(mark(X))
r12: mark(pairs()) -> a__pairs()
r13: mark(odds()) -> a__odds()
r14: mark(head(X)) -> a__head(mark(X))
r15: mark(tail(X)) -> a__tail(mark(X))
r16: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r17: mark(|0|()) -> |0|()
r18: mark(s(X)) -> s(mark(X))
r19: a__pairs() -> pairs()

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

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      a__odds#_A() = 3
      a__incr#_A(x1) = x1 + 1
      a__pairs_A() = 1
      cons_A(x1,x2) = x1
      mark#_A(x1) = x1
      incr_A(x1) = x1 + 1
      mark_A(x1) = x1
      odds_A() = 4
      s_A(x1) = x1
      a__nats_A() = 1
      |0|_A() = 1
      nats_A() = 1
      a__odds_A() = 4
      a__incr_A(x1) = x1 + 1
      a__head_A(x1) = x1 + 1
      head_A(x1) = x1 + 1
      a__tail_A(x1) = x1 + 1
      tail_A(x1) = x1 + 1
      pairs_A() = 1

The next rules are strictly ordered:

  p1, p2, p4, p5

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

and R consists of:

r1: a__nats() -> cons(|0|(),incr(nats()))
r2: a__odds() -> a__incr(a__pairs())
r3: a__incr(cons(X,XS)) -> cons(s(mark(X)),incr(XS))
r4: a__nats() -> nats()
r5: a__incr(X) -> incr(X)
r6: a__odds() -> odds()
r7: a__head(X) -> head(X)
r8: a__tail(X) -> tail(X)
r9: a__pairs() -> cons(|0|(),incr(odds()))
r10: mark(nats()) -> a__nats()
r11: mark(incr(X)) -> a__incr(mark(X))
r12: mark(pairs()) -> a__pairs()
r13: mark(odds()) -> a__odds()
r14: mark(head(X)) -> a__head(mark(X))
r15: mark(tail(X)) -> a__tail(mark(X))
r16: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r17: mark(|0|()) -> |0|()
r18: mark(s(X)) -> s(mark(X))
r19: a__pairs() -> pairs()

The estimated dependency graph contains the following SCCs:

  {p2, p3}


-- Reduction pair.

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

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

and R consists of:

r1: a__nats() -> cons(|0|(),incr(nats()))
r2: a__odds() -> a__incr(a__pairs())
r3: a__incr(cons(X,XS)) -> cons(s(mark(X)),incr(XS))
r4: a__nats() -> nats()
r5: a__incr(X) -> incr(X)
r6: a__odds() -> odds()
r7: a__head(X) -> head(X)
r8: a__tail(X) -> tail(X)
r9: a__pairs() -> cons(|0|(),incr(odds()))
r10: mark(nats()) -> a__nats()
r11: mark(incr(X)) -> a__incr(mark(X))
r12: mark(pairs()) -> a__pairs()
r13: mark(odds()) -> a__odds()
r14: mark(head(X)) -> a__head(mark(X))
r15: mark(tail(X)) -> a__tail(mark(X))
r16: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r17: mark(|0|()) -> |0|()
r18: mark(s(X)) -> s(mark(X))
r19: a__pairs() -> pairs()

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      mark#_A(x1) = x1
      cons_A(x1,x2) = x1 + 1
      s_A(x1) = x1

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: a__nats() -> cons(|0|(),incr(nats()))
r2: a__odds() -> a__incr(a__pairs())
r3: a__incr(cons(X,XS)) -> cons(s(mark(X)),incr(XS))
r4: a__nats() -> nats()
r5: a__incr(X) -> incr(X)
r6: a__odds() -> odds()
r7: a__head(X) -> head(X)
r8: a__tail(X) -> tail(X)
r9: a__pairs() -> cons(|0|(),incr(odds()))
r10: mark(nats()) -> a__nats()
r11: mark(incr(X)) -> a__incr(mark(X))
r12: mark(pairs()) -> a__pairs()
r13: mark(odds()) -> a__odds()
r14: mark(head(X)) -> a__head(mark(X))
r15: mark(tail(X)) -> a__tail(mark(X))
r16: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r17: mark(|0|()) -> |0|()
r18: mark(s(X)) -> s(mark(X))
r19: a__pairs() -> pairs()

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: a__nats() -> cons(|0|(),incr(nats()))
r2: a__odds() -> a__incr(a__pairs())
r3: a__incr(cons(X,XS)) -> cons(s(mark(X)),incr(XS))
r4: a__nats() -> nats()
r5: a__incr(X) -> incr(X)
r6: a__odds() -> odds()
r7: a__head(X) -> head(X)
r8: a__tail(X) -> tail(X)
r9: a__pairs() -> cons(|0|(),incr(odds()))
r10: mark(nats()) -> a__nats()
r11: mark(incr(X)) -> a__incr(mark(X))
r12: mark(pairs()) -> a__pairs()
r13: mark(odds()) -> a__odds()
r14: mark(head(X)) -> a__head(mark(X))
r15: mark(tail(X)) -> a__tail(mark(X))
r16: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r17: mark(|0|()) -> |0|()
r18: mark(s(X)) -> s(mark(X))
r19: a__pairs() -> pairs()

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      mark#_A(x1) = x1
      s_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19

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