YES

We show the termination of the TRS R:

  t(N) -> cs(r(q(N)),nt(ns(N)))
  q(|0|()) -> |0|()
  q(s(X)) -> s(p(q(X),d(X)))
  d(|0|()) -> |0|()
  d(s(X)) -> s(s(d(X)))
  p(|0|(),X) -> X
  p(X,|0|()) -> X
  p(s(X),s(Y)) -> s(s(p(X,Y)))
  f(|0|(),X) -> nil()
  f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
  t(X) -> nt(X)
  s(X) -> ns(X)
  f(X1,X2) -> nf(X1,X2)
  a(nt(X)) -> t(a(X))
  a(ns(X)) -> s(a(X))
  a(nf(X1,X2)) -> f(a(X1),a(X2))
  a(X) -> X

-- SCC decomposition.

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

p1: t#(N) -> q#(N)
p2: q#(s(X)) -> s#(p(q(X),d(X)))
p3: q#(s(X)) -> p#(q(X),d(X))
p4: q#(s(X)) -> q#(X)
p5: q#(s(X)) -> d#(X)
p6: d#(s(X)) -> s#(s(d(X)))
p7: d#(s(X)) -> s#(d(X))
p8: d#(s(X)) -> d#(X)
p9: p#(s(X),s(Y)) -> s#(s(p(X,Y)))
p10: p#(s(X),s(Y)) -> s#(p(X,Y))
p11: p#(s(X),s(Y)) -> p#(X,Y)
p12: f#(s(X),cs(Y,Z)) -> a#(Z)
p13: a#(nt(X)) -> t#(a(X))
p14: a#(nt(X)) -> a#(X)
p15: a#(ns(X)) -> s#(a(X))
p16: a#(ns(X)) -> a#(X)
p17: a#(nf(X1,X2)) -> f#(a(X1),a(X2))
p18: a#(nf(X1,X2)) -> a#(X1)
p19: a#(nf(X1,X2)) -> a#(X2)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

The estimated dependency graph contains the following SCCs:

  {p12, p14, p16, p17, p18, p19}
  {p4}
  {p11}
  {p8}


-- Reduction pair.

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

p1: a#(nf(X1,X2)) -> a#(X2)
p2: a#(nf(X1,X2)) -> a#(X1)
p3: a#(nf(X1,X2)) -> f#(a(X1),a(X2))
p4: f#(s(X),cs(Y,Z)) -> a#(Z)
p5: a#(ns(X)) -> a#(X)
p6: a#(nt(X)) -> a#(X)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> 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

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      a#_A(x1) = x1
      nf_A(x1,x2) = x1 + x2 + 2
      f#_A(x1,x2) = x2 + 1
      a_A(x1) = x1
      s_A(x1) = x1
      cs_A(x1,x2) = x2
      ns_A(x1) = x1
      nt_A(x1) = x1 + 1
      d_A(x1) = 0
      |0|_A() = 0
      p_A(x1,x2) = x1 + x2
      q_A(x1) = x1 + 1
      t_A(x1) = x1 + 1
      r_A(x1) = 1
      f_A(x1,x2) = x1 + x2 + 2
      nil_A() = 0

The next rules are strictly ordered:

  p1, p2, p3, p4, p6

We remove them from the problem.

-- SCC decomposition.

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

p1: a#(ns(X)) -> a#(X)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: a#(ns(X)) -> a#(X)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      a#_A(x1) = x1
      ns_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

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

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      q#_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

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: p#(s(X),s(Y)) -> p#(X,Y)

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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

and R consists of:

r1: t(N) -> cs(r(q(N)),nt(ns(N)))
r2: q(|0|()) -> |0|()
r3: q(s(X)) -> s(p(q(X),d(X)))
r4: d(|0|()) -> |0|()
r5: d(s(X)) -> s(s(d(X)))
r6: p(|0|(),X) -> X
r7: p(X,|0|()) -> X
r8: p(s(X),s(Y)) -> s(s(p(X,Y)))
r9: f(|0|(),X) -> nil()
r10: f(s(X),cs(Y,Z)) -> cs(Y,nf(X,a(Z)))
r11: t(X) -> nt(X)
r12: s(X) -> ns(X)
r13: f(X1,X2) -> nf(X1,X2)
r14: a(nt(X)) -> t(a(X))
r15: a(ns(X)) -> s(a(X))
r16: a(nf(X1,X2)) -> f(a(X1),a(X2))
r17: a(X) -> X

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      d#_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

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