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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              if_int#_A(x1,x2,x3,x4) = 0
              false_A() = 10
              if2#_A(x1,x2,x3) = 0
              int#_A(x1,x2) = 0
              p_A(x1) = 9
              zero_A(x1) = x1 + 1
              true_A() = 10
              if1#_A(x1,x2,x3) = 0
              s_A(x1) = 9
              |0|_A() = 9
              tail_A(x1) = 6
              nil_A() = 6
              if_int#!_A(x1,x2,x3,x4) = x1
              false!_A() = 9
              if2#!_A(x1,x2,x3) = 10
              int#!_A(x1,x2) = x1 + 1
              p!_A(x1) = 11
              zero!_A(x1) = 2
              true!_A() = 3
              if1#!_A(x1,x2,x3) = 10
              s!_A(x1) = x1 + 12
              |0|!_A() = 3
              tail!_A(x1) = 7
              nil!_A() = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            if_int#_A(x1,x2,x3,x4) = max{24, x1 - 16, x2 - 4, x3 - 8, x4 - 12}
            false_A = 16
            if2#_A(x1,x2,x3) = max{24, x1 - 21, x2 - 8, x3 - 12}
            int#_A(x1,x2) = max{24, x1 - 8, x2 - 12}
            p_A(x1) = max{8, x1 - 3}
            zero_A(x1) = max{5, x1 - 8}
            true_A = 0
            if1#_A(x1,x2,x3) = max{24, x1 - 20, x3 - 12}
            s_A(x1) = x1 + 24
            |0|_A = 8
            tail_A(x1) = max{30, x1 - 28}
            nil_A = 30
            if_int#!_A(x1,x2,x3,x4) = max{x1 + 27, x2 + 47, x3 + 19, x4 + 53}
            false!_A = 24
            if2#!_A(x1,x2,x3) = max{x1 + 46, x2 + 18, x3 + 51}
            int#!_A(x1,x2) = max{x1 + 20, x2 + 53}
            p!_A(x1) = 3
            zero!_A(x1) = 23
            true!_A = 27
            if1#!_A(x1,x2,x3) = max{x1 + 47, x3 + 53}
            s!_A(x1) = max{23, x1 - 1}
            |0|!_A = 27
            tail!_A(x1) = max{21, x1 - 30}
            nil!_A = 20
        
  
    partial status:
  
      pi(if_int#) = []
      pi(false) = []
      pi(if2#) = []
      pi(int#) = []
      pi(p) = [1]
      pi(zero) = [1]
      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

  r6, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              int#_A(x1,x2) = 0
              if_int#_A(x1,x2,x3,x4) = 0
              zero_A(x1) = 2
              true_A() = 2
              if1#_A(x1,x2,x3) = 0
              false_A() = 2
              s_A(x1) = 2
              |0|_A() = 6
              int#!_A(x1,x2) = x1 + 7
              if_int#!_A(x1,x2,x3,x4) = x1 + x2 + 2
              zero!_A(x1) = x1 + 6
              true!_A() = 11
              if1#!_A(x1,x2,x3) = 10
              false!_A() = 7
              s!_A(x1) = 14
              |0|!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            int#_A(x1,x2) = max{0, x1 - 8, x2 - 2}
            if_int#_A(x1,x2,x3,x4) = max{0, x1 - 5, x2 - 5, x3 - 8, x4 - 2}
            zero_A(x1) = max{4, x1 - 8}
            true_A = 4
            if1#_A(x1,x2,x3) = max{0, x1 - 5, x2 - 8, x3 - 2}
            false_A = 1
            s_A(x1) = max{7, x1}
            |0|_A = 7
            int#!_A(x1,x2) = x1 + 11
            if_int#!_A(x1,x2,x3,x4) = max{10, x1 - 5, x3 - 1}
            zero!_A(x1) = max{21, x1 - 8}
            true!_A = 20
            if1#!_A(x1,x2,x3) = 19
            false!_A = 16
            s!_A(x1) = max{72, x1 + 51}
            |0|!_A = 3
        
  
    partial status:
  
      pi(int#) = []
      pi(if_int#) = [1]
      pi(zero) = []
      pi(true) = []
      pi(if1#) = []
      pi(false) = []
      pi(s) = []
      pi(|0|) = []

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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              p#_A(x1) = x1
              s_A(x1) = 2
              p#!_A(x1) = x1 + 3
              s!_A(x1) = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            p#_A(x1) = x1
            s_A(x1) = max{31, x1 + 17}
            p#!_A(x1) = max{1, x1 - 29}
            s!_A(x1) = 0
        
  
    partial status:
  
      pi(p#) = []
      pi(s) = []

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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              intlist#_A(x1) = 0
              if_intlist#_A(x1,x2) = 0
              empty_A(x1) = x1 + 1
              false_A() = 4
              tail_A(x1) = x1 + 1
              nil_A() = 6
              true_A() = 7
              cons_A(x1,x2) = x2 + 3
              p_A(x1) = x1 + 1
              |0|_A() = 2
              intlist#!_A(x1) = 7
              if_intlist#!_A(x1,x2) = 7
              empty!_A(x1) = 6
              false!_A() = 5
              tail!_A(x1) = 0
              nil!_A() = 1
              true!_A() = 0
              cons!_A(x1,x2) = 6
              p!_A(x1) = 5
              |0|!_A() = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            intlist#_A(x1) = x1 + 3
            if_intlist#_A(x1,x2) = max{3, x2 + 1}
            empty_A(x1) = max{6, x1 - 40}
            false_A = 8
            tail_A(x1) = max{0, x1 - 2}
            nil_A = 0
            true_A = 6
            cons_A(x1,x2) = max{89, x1 + 26, x2 + 39}
            p_A(x1) = 26
            |0|_A = 4
            intlist#!_A(x1) = x1 + 52
            if_intlist#!_A(x1,x2) = max{x1 + 45, x2 + 51}
            empty!_A(x1) = max{48, x1 - 40}
            false!_A = 49
            tail!_A(x1) = 29
            nil!_A = 30
            true!_A = 29
            cons!_A(x1,x2) = max{50, x1 - 75, x2 - 4}
            p!_A(x1) = 49
            |0|!_A = 47
        
  
    partial status:
  
      pi(intlist#) = []
      pi(if_intlist#) = []
      pi(empty) = []
      pi(false) = []
      pi(tail) = [1]
      pi(nil) = []
      pi(true) = []
      pi(cons) = [1, 2]
      pi(p) = []
      pi(|0|) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.