YES

We show the termination of the TRS R:

  empty(nil()) -> true()
  empty(cons(x,y)) -> false()
  tail(nil()) -> nil()
  tail(cons(x,y)) -> y
  head(cons(x,y)) -> x
  zero(|0|()) -> true()
  zero(s(x)) -> false()
  p(|0|()) -> |0|()
  p(s(|0|())) -> |0|()
  p(s(s(x))) -> s(p(s(x)))
  intlist(x) -> if_intlist(empty(x),x)
  if_intlist(true(),x) -> nil()
  if_intlist(false(),x) -> cons(s(head(x)),intlist(tail(x)))
  int(x,y) -> if_int(zero(x),zero(y),x,y)
  if_int(true(),b,x,y) -> if1(b,x,y)
  if_int(false(),b,x,y) -> if2(b,x,y)
  if1(true(),x,y) -> cons(|0|(),nil())
  if1(false(),x,y) -> cons(|0|(),int(s(|0|()),y))
  if2(true(),x,y) -> nil()
  if2(false(),x,y) -> intlist(int(p(x),p(y)))

-- SCC decomposition.

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

p1: p#(s(s(x))) -> p#(s(x))
p2: intlist#(x) -> if_intlist#(empty(x),x)
p3: intlist#(x) -> empty#(x)
p4: if_intlist#(false(),x) -> head#(x)
p5: if_intlist#(false(),x) -> intlist#(tail(x))
p6: if_intlist#(false(),x) -> tail#(x)
p7: int#(x,y) -> if_int#(zero(x),zero(y),x,y)
p8: int#(x,y) -> zero#(x)
p9: int#(x,y) -> zero#(y)
p10: if_int#(true(),b,x,y) -> if1#(b,x,y)
p11: if_int#(false(),b,x,y) -> if2#(b,x,y)
p12: if1#(false(),x,y) -> int#(s(|0|()),y)
p13: if2#(false(),x,y) -> intlist#(int(p(x),p(y)))
p14: if2#(false(),x,y) -> int#(p(x),p(y))
p15: if2#(false(),x,y) -> p#(x)
p16: if2#(false(),x,y) -> p#(y)

and R consists of:

r1: empty(nil()) -> true()
r2: empty(cons(x,y)) -> false()
r3: tail(nil()) -> nil()
r4: tail(cons(x,y)) -> y
r5: head(cons(x,y)) -> x
r6: zero(|0|()) -> true()
r7: zero(s(x)) -> false()
r8: p(|0|()) -> |0|()
r9: p(s(|0|())) -> |0|()
r10: p(s(s(x))) -> s(p(s(x)))
r11: intlist(x) -> if_intlist(empty(x),x)
r12: if_intlist(true(),x) -> nil()
r13: if_intlist(false(),x) -> cons(s(head(x)),intlist(tail(x)))
r14: int(x,y) -> if_int(zero(x),zero(y),x,y)
r15: if_int(true(),b,x,y) -> if1(b,x,y)
r16: if_int(false(),b,x,y) -> if2(b,x,y)
r17: if1(true(),x,y) -> cons(|0|(),nil())
r18: if1(false(),x,y) -> cons(|0|(),int(s(|0|()),y))
r19: if2(true(),x,y) -> nil()
r20: if2(false(),x,y) -> intlist(int(p(x),p(y)))

The estimated dependency graph contains the following SCCs:

  {p7, p10, p11, p12, p14}
  {p1}
  {p2, p5}


-- Reduction pair.

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

p1: if_int#(false(),b,x,y) -> if2#(b,x,y)
p2: if2#(false(),x,y) -> int#(p(x),p(y))
p3: int#(x,y) -> if_int#(zero(x),zero(y),x,y)
p4: if_int#(true(),b,x,y) -> if1#(b,x,y)
p5: if1#(false(),x,y) -> int#(s(|0|()),y)

and R consists of:

r1: empty(nil()) -> true()
r2: empty(cons(x,y)) -> false()
r3: tail(nil()) -> nil()
r4: tail(cons(x,y)) -> y
r5: head(cons(x,y)) -> x
r6: zero(|0|()) -> true()
r7: zero(s(x)) -> false()
r8: p(|0|()) -> |0|()
r9: p(s(|0|())) -> |0|()
r10: p(s(s(x))) -> s(p(s(x)))
r11: intlist(x) -> if_intlist(empty(x),x)
r12: if_intlist(true(),x) -> nil()
r13: if_intlist(false(),x) -> cons(s(head(x)),intlist(tail(x)))
r14: int(x,y) -> if_int(zero(x),zero(y),x,y)
r15: if_int(true(),b,x,y) -> if1(b,x,y)
r16: if_int(false(),b,x,y) -> if2(b,x,y)
r17: if1(true(),x,y) -> cons(|0|(),nil())
r18: if1(false(),x,y) -> cons(|0|(),int(s(|0|()),y))
r19: if2(true(),x,y) -> nil()
r20: if2(false(),x,y) -> intlist(int(p(x),p(y)))

The set of usable rules consists of

  r6, r7, r8, r9, r10

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    if_int#_A(x1,x2,x3,x4) = max{x1 + 1, x2 + 7, x3 + 2, x4 + 9}
    false_A = 10
    if2#_A(x1,x2,x3) = max{10, x1 + 7, x2 + 1, x3 + 9}
    int#_A(x1,x2) = max{x1 + 3, x2 + 9}
    p_A(x1) = max{0, x1 - 3}
    zero_A(x1) = x1 + 1
    true_A = 1
    if1#_A(x1,x2,x3) = max{x1 + 6, x2 + 1, x3 + 9}
    s_A(x1) = x1 + 9
    |0|_A = 0

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: if_int#(false(),b,x,y) -> if2#(b,x,y)
p2: int#(x,y) -> if_int#(zero(x),zero(y),x,y)
p3: if_int#(true(),b,x,y) -> if1#(b,x,y)
p4: if1#(false(),x,y) -> int#(s(|0|()),y)

and R consists of:

r1: empty(nil()) -> true()
r2: empty(cons(x,y)) -> false()
r3: tail(nil()) -> nil()
r4: tail(cons(x,y)) -> y
r5: head(cons(x,y)) -> x
r6: zero(|0|()) -> true()
r7: zero(s(x)) -> false()
r8: p(|0|()) -> |0|()
r9: p(s(|0|())) -> |0|()
r10: p(s(s(x))) -> s(p(s(x)))
r11: intlist(x) -> if_intlist(empty(x),x)
r12: if_intlist(true(),x) -> nil()
r13: if_intlist(false(),x) -> cons(s(head(x)),intlist(tail(x)))
r14: int(x,y) -> if_int(zero(x),zero(y),x,y)
r15: if_int(true(),b,x,y) -> if1(b,x,y)
r16: if_int(false(),b,x,y) -> if2(b,x,y)
r17: if1(true(),x,y) -> cons(|0|(),nil())
r18: if1(false(),x,y) -> cons(|0|(),int(s(|0|()),y))
r19: if2(true(),x,y) -> nil()
r20: if2(false(),x,y) -> intlist(int(p(x),p(y)))

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4}


-- Reduction pair.

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

p1: int#(x,y) -> if_int#(zero(x),zero(y),x,y)
p2: if_int#(true(),b,x,y) -> if1#(b,x,y)
p3: if1#(false(),x,y) -> int#(s(|0|()),y)

and R consists of:

r1: empty(nil()) -> true()
r2: empty(cons(x,y)) -> false()
r3: tail(nil()) -> nil()
r4: tail(cons(x,y)) -> y
r5: head(cons(x,y)) -> x
r6: zero(|0|()) -> true()
r7: zero(s(x)) -> false()
r8: p(|0|()) -> |0|()
r9: p(s(|0|())) -> |0|()
r10: p(s(s(x))) -> s(p(s(x)))
r11: intlist(x) -> if_intlist(empty(x),x)
r12: if_intlist(true(),x) -> nil()
r13: if_intlist(false(),x) -> cons(s(head(x)),intlist(tail(x)))
r14: int(x,y) -> if_int(zero(x),zero(y),x,y)
r15: if_int(true(),b,x,y) -> if1(b,x,y)
r16: if_int(false(),b,x,y) -> if2(b,x,y)
r17: if1(true(),x,y) -> cons(|0|(),nil())
r18: if1(false(),x,y) -> cons(|0|(),int(s(|0|()),y))
r19: if2(true(),x,y) -> nil()
r20: if2(false(),x,y) -> intlist(int(p(x),p(y)))

The set of usable rules consists of

  r6, r7

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    int#_A(x1,x2) = x1 + 5
    if_int#_A(x1,x2,x3,x4) = max{x1 - 1, x3 + 1}
    zero_A(x1) = x1 + 5
    true_A = 7
    if1#_A(x1,x2,x3) = 6
    false_A = 1
    s_A(x1) = max{0, x1 - 4}
    |0|_A = 2

The next rules are strictly ordered:

  p1, p3

We remove them from the problem.

-- SCC decomposition.

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

p1: if_int#(true(),b,x,y) -> if1#(b,x,y)

and R consists of:

r1: empty(nil()) -> true()
r2: empty(cons(x,y)) -> false()
r3: tail(nil()) -> nil()
r4: tail(cons(x,y)) -> y
r5: head(cons(x,y)) -> x
r6: zero(|0|()) -> true()
r7: zero(s(x)) -> false()
r8: p(|0|()) -> |0|()
r9: p(s(|0|())) -> |0|()
r10: p(s(s(x))) -> s(p(s(x)))
r11: intlist(x) -> if_intlist(empty(x),x)
r12: if_intlist(true(),x) -> nil()
r13: if_intlist(false(),x) -> cons(s(head(x)),intlist(tail(x)))
r14: int(x,y) -> if_int(zero(x),zero(y),x,y)
r15: if_int(true(),b,x,y) -> if1(b,x,y)
r16: if_int(false(),b,x,y) -> if2(b,x,y)
r17: if1(true(),x,y) -> cons(|0|(),nil())
r18: if1(false(),x,y) -> cons(|0|(),int(s(|0|()),y))
r19: if2(true(),x,y) -> nil()
r20: if2(false(),x,y) -> intlist(int(p(x),p(y)))

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

p1: p#(s(s(x))) -> p#(s(x))

and R consists of:

r1: empty(nil()) -> true()
r2: empty(cons(x,y)) -> false()
r3: tail(nil()) -> nil()
r4: tail(cons(x,y)) -> y
r5: head(cons(x,y)) -> x
r6: zero(|0|()) -> true()
r7: zero(s(x)) -> false()
r8: p(|0|()) -> |0|()
r9: p(s(|0|())) -> |0|()
r10: p(s(s(x))) -> s(p(s(x)))
r11: intlist(x) -> if_intlist(empty(x),x)
r12: if_intlist(true(),x) -> nil()
r13: if_intlist(false(),x) -> cons(s(head(x)),intlist(tail(x)))
r14: int(x,y) -> if_int(zero(x),zero(y),x,y)
r15: if_int(true(),b,x,y) -> if1(b,x,y)
r16: if_int(false(),b,x,y) -> if2(b,x,y)
r17: if1(true(),x,y) -> cons(|0|(),nil())
r18: if1(false(),x,y) -> cons(|0|(),int(s(|0|()),y))
r19: if2(true(),x,y) -> nil()
r20: if2(false(),x,y) -> intlist(int(p(x),p(y)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    p#_A(x1) = max{0, x1 - 1}
    s_A(x1) = x1 + 2

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: intlist#(x) -> if_intlist#(empty(x),x)
p2: if_intlist#(false(),x) -> intlist#(tail(x))

and R consists of:

r1: empty(nil()) -> true()
r2: empty(cons(x,y)) -> false()
r3: tail(nil()) -> nil()
r4: tail(cons(x,y)) -> y
r5: head(cons(x,y)) -> x
r6: zero(|0|()) -> true()
r7: zero(s(x)) -> false()
r8: p(|0|()) -> |0|()
r9: p(s(|0|())) -> |0|()
r10: p(s(s(x))) -> s(p(s(x)))
r11: intlist(x) -> if_intlist(empty(x),x)
r12: if_intlist(true(),x) -> nil()
r13: if_intlist(false(),x) -> cons(s(head(x)),intlist(tail(x)))
r14: int(x,y) -> if_int(zero(x),zero(y),x,y)
r15: if_int(true(),b,x,y) -> if1(b,x,y)
r16: if_int(false(),b,x,y) -> if2(b,x,y)
r17: if1(true(),x,y) -> cons(|0|(),nil())
r18: if1(false(),x,y) -> cons(|0|(),int(s(|0|()),y))
r19: if2(true(),x,y) -> nil()
r20: if2(false(),x,y) -> intlist(int(p(x),p(y)))

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    intlist#_A(x1) = x1 + 2
    if_intlist#_A(x1,x2) = max{x1 + 2, x2 - 1}
    empty_A(x1) = x1
    false_A = 2
    tail_A(x1) = max{1, x1 - 4}
    nil_A = 1
    true_A = 1
    cons_A(x1,x2) = max{x1 + 2, x2 + 4}

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: intlist#(x) -> if_intlist#(empty(x),x)

and R consists of:

r1: empty(nil()) -> true()
r2: empty(cons(x,y)) -> false()
r3: tail(nil()) -> nil()
r4: tail(cons(x,y)) -> y
r5: head(cons(x,y)) -> x
r6: zero(|0|()) -> true()
r7: zero(s(x)) -> false()
r8: p(|0|()) -> |0|()
r9: p(s(|0|())) -> |0|()
r10: p(s(s(x))) -> s(p(s(x)))
r11: intlist(x) -> if_intlist(empty(x),x)
r12: if_intlist(true(),x) -> nil()
r13: if_intlist(false(),x) -> cons(s(head(x)),intlist(tail(x)))
r14: int(x,y) -> if_int(zero(x),zero(y),x,y)
r15: if_int(true(),b,x,y) -> if1(b,x,y)
r16: if_int(false(),b,x,y) -> if2(b,x,y)
r17: if1(true(),x,y) -> cons(|0|(),nil())
r18: if1(false(),x,y) -> cons(|0|(),int(s(|0|()),y))
r19: if2(true(),x,y) -> nil()
r20: if2(false(),x,y) -> intlist(int(p(x),p(y)))

The estimated dependency graph contains the following SCCs:

  (no SCCs)