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

  r2, r3, r6, r7, r8, r9, r10, r12, r15, r17, r18, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_int#_A(x1,x2,x3,x4) = max{x2 - 46, x3 + 27, x4 + 68}
        false_A = 141
        if2#_A(x1,x2,x3) = max{x1 - 47, x2 + 23, x3 + 58}
        int#_A(x1,x2) = max{x1 + 27, x2 + 68}
        p_A(x1) = max{25, x1 - 11}
        zero_A(x1) = x1 + 106
        true_A = 72
        if1#_A(x1,x2,x3) = max{x1 - 125, x3 + 68}
        s_A(x1) = max{36, x1 + 24}
        |0|_A = 17
        empty_A(x1) = 142
        cons_A(x1,x2) = max{2, x1 - 170}
        tail_A(x1) = max{144, x1 - 174}
        nil_A = 71
        if_intlist_A(x1,x2) = max{145, x1 - 170, x2 - 173}
        if_int_A(x1,x2,x3,x4) = max{172, x1 - 71, x2 - 107, x3 - 151, x4 - 1}
        if1_A(x1,x2,x3) = max{2, x1 - 295, x2 - 152, x3 - 2}
        int_A(x1,x2) = max{171, x1 - 39, x2 - 3}
        if2_A(x1,x2,x3) = max{172, x1 - 328, x2 - 151, x3 - 10}
  
    base order 2:
  
      precedence:
      
        if_int > if_intlist = if2 > if2# = tail > p = empty > if_int# = false = int# = zero = if1# = if1 > true = s = int > |0| = cons = nil
      
  
    partial status:
  
      pi(if_int#) = [4]
      pi(false) = []
      pi(if2#) = []
      pi(int#) = [2]
      pi(p) = []
      pi(zero) = []
      pi(true) = []
      pi(if1#) = [3]
      pi(s) = []
      pi(|0|) = []
      pi(empty) = []
      pi(cons) = []
      pi(tail) = []
      pi(nil) = []
      pi(if_intlist) = []
      pi(if_int) = []
      pi(if1) = []
      pi(int) = []
      pi(if2) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: if2#(false(),x,y) -> int#(p(x),p(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, r8, r10, r12, r17, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        int#_A(x1,x2) = max{4, x1 + 2}
        if_int#_A(x1,x2,x3,x4) = max{3, x1 - 10, x3 + 1}
        zero_A(x1) = x1 + 9
        true_A = 16
        if1#_A(x1,x2,x3) = 5
        false_A = 8
        s_A(x1) = 0
        |0|_A = 7
        tail_A(x1) = max{54, x1 - 2}
        nil_A = 1
        p_A(x1) = max{35, x1 - 28}
        if_intlist_A(x1,x2) = max{10, x1 - 54}
        if1_A(x1,x2,x3) = max{17, x1 - 45, x2 - 63, x3 - 42}
        cons_A(x1,x2) = max{9, x1 - 45, x2}
        if2_A(x1,x2,x3) = max{3, x1 - 72, x2 - 64, x3 - 71}
  
    base order 2:
  
      precedence:
      
        p > tail = if1 = if2 > cons > zero = if_intlist > true = false > int# = if_int# = nil > if1# > s > |0|
      
  
    partial status:
  
      pi(int#) = [1]
      pi(if_int#) = []
      pi(zero) = [1]
      pi(true) = []
      pi(if1#) = []
      pi(false) = []
      pi(s) = []
      pi(|0|) = []
      pi(tail) = []
      pi(nil) = []
      pi(p) = []
      pi(if_intlist) = []
      pi(if1) = []
      pi(cons) = [2]
      pi(if2) = []

The following rules are removed:

  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: 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:

  (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

  r2, r3, r7, r9, r10, r12, r17, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        p#_A(x1) = x1 + 16
        s_A(x1) = x1 + 17
        empty_A(x1) = max{4, x1 - 10}
        cons_A(x1,x2) = max{2, x1 - 30, x2 - 6}
        false_A = 1
        tail_A(x1) = max{8, x1 - 11}
        nil_A = 0
        zero_A(x1) = max{7, x1 - 21}
        p_A(x1) = max{3, x1 - 14}
        |0|_A = 2
        if_intlist_A(x1,x2) = max{19, x1 - 19, x2 - 10}
        true_A = 3
        if1_A(x1,x2,x3) = max{20, x1 - 32, x2 - 14, x3 - 14}
        if2_A(x1,x2,x3) = max{9, x1 - 23, x2 - 13, x3 - 13}
  
    base order 2:
  
      precedence:
      
        if_intlist = if1 = if2 > tail = p > p# = s = empty = zero > true > cons = nil = |0| > false
      
  
    partial status:
  
      pi(p#) = [1]
      pi(s) = [1]
      pi(empty) = []
      pi(cons) = []
      pi(false) = []
      pi(tail) = []
      pi(nil) = []
      pi(zero) = []
      pi(p) = []
      pi(|0|) = []
      pi(if_intlist) = []
      pi(true) = []
      pi(if1) = []
      pi(if2) = []

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, r17, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        intlist#_A(x1) = x1 + 16
        if_intlist#_A(x1,x2) = max{x1 - 12, x2 + 15}
        empty_A(x1) = max{15, x1 + 11}
        false_A = 54
        tail_A(x1) = max{22, x1 - 17}
        nil_A = 18
        true_A = 19
        cons_A(x1,x2) = max{54, x1 - 10, x2 + 41}
        zero_A(x1) = max{55, x1 - 95}
        s_A(x1) = x1 + 53
        p_A(x1) = max{119, x1 - 107}
        |0|_A = 7
        if_intlist_A(x1,x2) = max{99, x1 - 33, x2 + 13}
        if1_A(x1,x2,x3) = max{60, x1 - 148, x2 - 2, x3 - 53}
        if2_A(x1,x2,x3) = max{53, x1 - 75, x2 - 108, x3 - 108}
  
    base order 2:
  
      precedence:
      
        zero = if1 > true = s = p = if_intlist > intlist# = cons > if_intlist# = false > empty = tail = |0| = if2 > nil
      
  
    partial status:
  
      pi(intlist#) = [1]
      pi(if_intlist#) = []
      pi(empty) = [1]
      pi(false) = []
      pi(tail) = []
      pi(nil) = []
      pi(true) = []
      pi(cons) = []
      pi(zero) = []
      pi(s) = []
      pi(p) = []
      pi(|0|) = []
      pi(if_intlist) = [2]
      pi(if1) = []
      pi(if2) = []

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)