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

  r3, r6, r7, r8, r9, r10

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          if_int#_A(x1,x2,x3,x4) = x2 + 1
          false_A() = 4
          if2#_A(x1,x2,x3) = x1 + 1
          int#_A(x1,x2) = 5
          p_A(x1) = 7
          zero_A(x1) = 4
          true_A() = 2
          if1#_A(x1,x2,x3) = x1 + 1
          s_A(x1) = 6
          |0|_A() = 3
          tail_A(x1) = 0
          nil_A() = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_int#_A(x1,x2,x3,x4) = max{37, x1 - 81, x2 - 26, x3 + 1, x4 + 36}
          false_A = 117
          if2#_A(x1,x2,x3) = max{37, x1 - 60, x2 - 1, x3 + 18}
          int#_A(x1,x2) = max{56, x1 + 17, x2 + 36}
          p_A(x1) = max{21, x1 - 18}
          zero_A(x1) = max{82, x1 + 62}
          true_A = 82
          if1#_A(x1,x2,x3) = max{x1 - 26, x2 + 1, x3 + 36}
          s_A(x1) = x1 + 55
          |0|_A = 19
          tail_A(x1) = 191
          nil_A = 5
          if_int#!_A(x1,x2,x3,x4) = max{x2 + 11, x3 + 19, x4 + 94}
          false!_A = 7
          if2#!_A(x1,x2,x3) = max{x1 - 1, x2 + 19, x3 + 94}
          int#!_A(x1,x2) = max{x1 + 19, x2 + 94}
          p!_A(x1) = 1
          zero!_A(x1) = x1 + 95
          true!_A = 115
          if1#!_A(x1,x2,x3) = max{x2 + 19, x3 + 94}
          s!_A(x1) = 4
          |0|!_A = 116
          tail!_A(x1) = max{3, x1 - 6}
          nil!_A = 2
  
    partial status:
  
      pi(if_int#) = []
      pi(false) = []
      pi(if2#) = []
      pi(int#) = []
      pi(p) = []
      pi(zero) = []
      pi(true) = []
      pi(if1#) = []
      pi(s) = []
      pi(|0|) = []
      pi(tail) = []
      pi(nil) = []

The following rules are removed:

  p2



-- 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

  r3, r6, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          int#_A(x1,x2) = x1 + x2 + 2
          if_int#_A(x1,x2,x3,x4) = x1 + x4 + 1
          zero_A(x1) = x1 + 1
          true_A() = 6
          if1#_A(x1,x2,x3) = x3 + 5
          false_A() = 3
          s_A(x1) = 2
          |0|_A() = 7
          tail_A(x1) = 8
          nil_A() = 7
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          int#_A(x1,x2) = max{9, x1 + 4, x2 - 1}
          if_int#_A(x1,x2,x3,x4) = max{9, x3 - 1, x4 - 1}
          zero_A(x1) = x1 + 4
          true_A = 5
          if1#_A(x1,x2,x3) = max{9, x2 - 1, x3 - 1}
          false_A = 2
          s_A(x1) = max{5, x1 - 4}
          |0|_A = 1
          tail_A(x1) = max{23, x1 - 4}
          nil_A = 3
          int#!_A(x1,x2) = max{x1 + 2, x2 + 6}
          if_int#!_A(x1,x2,x3,x4) = max{5, x1 - 4, x2 + 1}
          zero!_A(x1) = max{6, x1 - 6}
          true!_A = 6
          if1#!_A(x1,x2,x3) = max{3, x2 + 1}
          false!_A = 6
          s!_A(x1) = x1 + 15
          |0|!_A = 7
          tail!_A(x1) = max{7, x1 - 4}
          nil!_A = 8
  
    partial status:
  
      pi(int#) = []
      pi(if_int#) = []
      pi(zero) = []
      pi(true) = []
      pi(if1#) = []
      pi(false) = []
      pi(s) = []
      pi(|0|) = []
      pi(tail) = []
      pi(nil) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.

-- 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

  r7, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          p#_A(x1) = x1 + 1
          s_A(x1) = x1 + 10
          zero_A(x1) = 6
          false_A() = 4
          p_A(x1) = 9
          |0|_A() = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          p#_A(x1) = 0
          s_A(x1) = max{0, x1 - 20}
          zero_A(x1) = max{44, x1 - 3}
          false_A = 9
          p_A(x1) = max{19, x1 - 30}
          |0|_A = 2
          p#!_A(x1) = 0
          s!_A(x1) = 0
          zero!_A(x1) = max{6, x1 - 3}
          false!_A = 5
          p!_A(x1) = x1 + 71
          |0|!_A = 72
  
    partial status:
  
      pi(p#) = [1]
      pi(s) = [1]
      pi(zero) = []
      pi(false) = []
      pi(p) = []
      pi(|0|) = []

The following rules are removed:

  p1

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, r7, r8, r12, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          intlist#_A(x1) = x1 + 12
          if_intlist#_A(x1,x2) = x2 + 12
          empty_A(x1) = 11
          false_A() = 10
          tail_A(x1) = x1
          nil_A() = 1
          true_A() = 5
          cons_A(x1,x2) = x2 + 5
          zero_A(x1) = 11
          s_A(x1) = x1 + 11
          p_A(x1) = 9
          |0|_A() = 8
          if_intlist_A(x1,x2) = 18
          if2_A(x1,x2,x3) = x2 + 13
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          intlist#_A(x1) = max{0, x1 - 1}
          if_intlist#_A(x1,x2) = max{0, x1 - 14, x2 - 2}
          empty_A(x1) = max{0, x1 - 69}
          false_A = 12
          tail_A(x1) = max{0, x1 - 68}
          nil_A = 0
          true_A = 0
          cons_A(x1,x2) = max{81, x1 - 1, x2 + 68}
          zero_A(x1) = max{34, x1 - 1}
          s_A(x1) = max{33, x1 - 33}
          p_A(x1) = max{32, x1 - 66}
          |0|_A = 0
          if_intlist_A(x1,x2) = max{52, x1 - 15, x2 - 68}
          if2_A(x1,x2,x3) = max{2, x1 - 134, x2 - 185, x3 - 220}
          intlist#!_A(x1) = max{4, x1 + 2}
          if_intlist#!_A(x1,x2) = max{x1 - 1, x2 + 2}
          empty!_A(x1) = max{3, x1 - 82}
          false!_A = 9
          tail!_A(x1) = max{10, x1 - 89}
          nil!_A = 6
          true!_A = 7
          cons!_A(x1,x2) = max{8, x1 - 23, x2 - 4}
          zero!_A(x1) = max{12, x1 - 34}
          s!_A(x1) = 10
          p!_A(x1) = 0
          |0|!_A = 11
          if_intlist!_A(x1,x2) = max{7, x1 - 35, x2 - 89}
          if2!_A(x1,x2,x3) = max{11, x1 - 13, x2 - 1, x3 + 5}
  
    partial status:
  
      pi(intlist#) = []
      pi(if_intlist#) = []
      pi(empty) = []
      pi(false) = []
      pi(tail) = []
      pi(nil) = []
      pi(true) = []
      pi(cons) = [2]
      pi(zero) = []
      pi(s) = [1]
      pi(p) = []
      pi(|0|) = []
      pi(if_intlist) = []
      pi(if2) = [2]

The following rules are removed:

  p2



-- 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)