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:
  
      max/plus interpretation on natural numbers:
      
        if_int#_A(x1,x2,x3,x4) = max{x1 + 2, x2 + 44, x3 + 27, x4 + 42}
        false_A = 88
        if2#_A(x1,x2,x3) = max{x1 - 1, x2 + 18, x3 + 34}
        int#_A(x1,x2) = max{86, x1 + 27, x2 + 42}
        p_A(x1) = max{17, x1 - 9}
        zero_A(x1) = max{41, x1 - 3}
        true_A = 14
        if1#_A(x1,x2,x3) = max{44, x1 + 43, x2 + 19, x3 + 42}
        s_A(x1) = x1 + 93
        |0|_A = 10
        tail_A(x1) = max{87, x1 - 13}
        nil_A = 12
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_int#_A(x1,x2,x3,x4) = 1
          false_A = 12
          if2#_A(x1,x2,x3) = 1
          int#_A(x1,x2) = 1
          p_A(x1) = max{14, x1 + 3}
          zero_A(x1) = max{9, x1 - 31}
          true_A = 7
          if1#_A(x1,x2,x3) = 1
          s_A(x1) = x1 + 43
          |0|_A = 37
          tail_A(x1) = max{1, x1}
          nil_A = 18
          if_int#!_A(x1,x2,x3,x4) = max{8, x4 - 4}
          false!_A = 21
          if2#!_A(x1,x2,x3) = max{x1 - 1, x2 + 3, x3 + 9}
          int#!_A(x1,x2) = max{8, x1 - 72, x2 - 4}
          p!_A(x1) = max{42, x1 + 2}
          zero!_A(x1) = max{7, x1 + 5}
          true!_A = 41
          if1#!_A(x1,x2,x3) = max{8, x3 - 4}
          s!_A(x1) = 41
          |0|!_A = 40
          tail!_A(x1) = max{0, x1 - 19}
          nil!_A = 0
  
    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:

  p1, p2



-- SCC decomposition.

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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- 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, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        int#_A(x1,x2) = max{6, x1 - 6}
        if_int#_A(x1,x2,x3,x4) = max{1, x1 - 50, x3 - 7}
        zero_A(x1) = x1 + 43
        true_A = 56
        if1#_A(x1,x2,x3) = 6
        false_A = 47
        s_A(x1) = max{5, x1 - 42}
        |0|_A = 14
        tail_A(x1) = max{59, x1 - 14}
        nil_A = 57
        p_A(x1) = max{15, x1 - 90}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          int#_A(x1,x2) = max{0, x1 - 208, x2 - 25}
          if_int#_A(x1,x2,x3,x4) = max{0, x1 - 65, x2 - 65, x3 - 208, x4 - 25}
          zero_A(x1) = max{41, x1 - 209}
          true_A = 39
          if1#_A(x1,x2,x3) = max{0, x1 - 65, x2 - 208, x3 - 25}
          false_A = 32
          s_A(x1) = max{184, x1 - 2}
          |0|_A = 209
          tail_A(x1) = max{109, x1 - 6}
          nil_A = 5
          p_A(x1) = max{209, x1 - 210}
          int#!_A(x1,x2) = max{43, x1 - 162}
          if_int#!_A(x1,x2,x3,x4) = max{48, x2 + 1, x3 + 45}
          zero!_A(x1) = max{1, x1 - 422}
          true!_A = 46
          if1#!_A(x1,x2,x3) = max{47, x1 - 1, x2 + 44}
          false!_A = 27
          s!_A(x1) = max{30, x1 - 211}
          |0|!_A = 45
          tail!_A(x1) = max{28, x1 - 81}
          nil!_A = 47
          p!_A(x1) = max{1, x1 - 210}
  
    partial status:
  
      pi(int#) = []
      pi(if_int#) = []
      pi(zero) = []
      pi(true) = []
      pi(if1#) = []
      pi(false) = []
      pi(s) = []
      pi(|0|) = []
      pi(tail) = []
      pi(nil) = []
      pi(p) = []

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

  r2, r7, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        p#_A(x1) = max{22, x1}
        s_A(x1) = x1 + 11
        empty_A(x1) = max{24, x1 - 24}
        cons_A(x1,x2) = max{13, x1 - 12, x2 - 32}
        false_A = 12
        zero_A(x1) = max{15, x1 - 15}
        p_A(x1) = max{9, x1 - 9}
        |0|_A = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          p#_A(x1) = 0
          s_A(x1) = max{9, x1 - 15}
          empty_A(x1) = max{0, x1 - 28}
          cons_A(x1,x2) = max{32, x1 - 25, x2 - 6}
          false_A = 4
          zero_A(x1) = max{14, x1 - 25}
          p_A(x1) = max{14, x1 + 4}
          |0|_A = 24
          p#!_A(x1) = x1 + 32
          s!_A(x1) = max{34, x1 + 33}
          empty!_A(x1) = max{59, x1 - 33}
          cons!_A(x1,x2) = max{1, x1 - 37, x2 - 6}
          false!_A = 58
          zero!_A(x1) = max{59, x1 - 42}
          p!_A(x1) = max{32, x1 - 29}
          |0|!_A = 30
  
    partial status:
  
      pi(p#) = [1]
      pi(s) = []
      pi(empty) = []
      pi(cons) = []
      pi(false) = []
      pi(zero) = []
      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, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        intlist#_A(x1) = 5
        if_intlist#_A(x1,x2) = 5
        empty_A(x1) = x1 + 44
        false_A = 9
        tail_A(x1) = x1 + 37
        nil_A = 14
        true_A = 12
        cons_A(x1,x2) = max{11, x1 - 43, x2 - 36}
        p_A(x1) = max{16, x1 - 46}
        |0|_A = 11
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          intlist#_A(x1) = max{7, x1 - 5}
          if_intlist#_A(x1,x2) = max{7, x1 - 2, x2 - 5}
          empty_A(x1) = max{1, x1 - 47}
          false_A = 8
          tail_A(x1) = max{4, x1 - 5}
          nil_A = 4
          true_A = 1
          cons_A(x1,x2) = max{55, x1 - 30, x2 + 46}
          p_A(x1) = 62
          |0|_A = 29
          intlist#!_A(x1) = max{4, x1 - 2}
          if_intlist#!_A(x1,x2) = max{1, x1 - 4, x2 - 7}
          empty!_A(x1) = max{10, x1 - 56}
          false!_A = 3
          tail!_A(x1) = max{8, x1 - 12}
          nil!_A = 10
          true!_A = 7
          cons!_A(x1,x2) = max{8, x1 - 66, x2 - 50}
          p!_A(x1) = max{5, x1 - 130}
          |0|!_A = 6
  
    partial status:
  
      pi(intlist#) = []
      pi(if_intlist#) = []
      pi(empty) = []
      pi(false) = []
      pi(tail) = []
      pi(nil) = []
      pi(true) = []
      pi(cons) = []
      pi(p) = []
      pi(|0|) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  (no SCCs)