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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_int#_A(x1,x2,x3,x4) = max{x2 + 46, x3 + 23, x4 + 46}
          false_A = 16
          if2#_A(x1,x2,x3) = max{x1 + 46, x2 + 22, x3 + 46}
          int#_A(x1,x2) = max{x1 + 24, x2 + 46}
          p_A(x1) = max{15, x1 - 3}
          zero_A(x1) = x1
          true_A = 0
          if1#_A(x1,x2,x3) = max{x1 + 1, x2 + 22, x3 + 46}
          s_A(x1) = max{21, x1 + 7}
          |0|_A = 0
    
      precedence:
      
        p > if_int# = false = int# = zero = true = if1# > if2# > s = |0|
      
      partial status:
    
        pi(if_int#) = [4]
        pi(false) = []
        pi(if2#) = [1]
        pi(int#) = [2]
        pi(p) = []
        pi(zero) = [1]
        pi(true) = []
        pi(if1#) = [3]
        pi(s) = [1]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if_int#_A(x1,x2,x3,x4) = x4 + 19
          false_A = 13
          if2#_A(x1,x2,x3) = 4
          int#_A(x1,x2) = x2 + 19
          p_A(x1) = 4
          zero_A(x1) = 13
          true_A = 12
          if1#_A(x1,x2,x3) = x3 + 19
          s_A(x1) = x1 + 5
          |0|_A = 8
    
      precedence:
      
        if_int# = false = int# = p = true = if1# > if2# = s > zero = |0|
      
      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|) = []
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.

-- 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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          int#_A(x1,x2) = x1 + 9
          if_int#_A(x1,x2,x3,x4) = max{8, x1 + 1, x3 + 2}
          zero_A(x1) = x1 + 8
          true_A = 16
          if1#_A(x1,x2,x3) = max{16, x2 + 1}
          false_A = 5
          s_A(x1) = max{6, x1 - 2}
          |0|_A = 9
    
      precedence:
      
        zero = true > false = s > if1# > |0| > int# = if_int#
      
      partial status:
    
        pi(int#) = []
        pi(if_int#) = []
        pi(zero) = [1]
        pi(true) = []
        pi(if1#) = []
        pi(false) = []
        pi(s) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          int#_A(x1,x2) = 10
          if_int#_A(x1,x2,x3,x4) = 10
          zero_A(x1) = max{7, x1 + 2}
          true_A = 2
          if1#_A(x1,x2,x3) = 7
          false_A = 8
          s_A(x1) = 7
          |0|_A = 1
    
      precedence:
      
        zero = false > s > int# = true > if_int# = if1# > |0|
      
      partial status:
    
        pi(int#) = []
        pi(if_int#) = []
        pi(zero) = []
        pi(true) = []
        pi(if1#) = []
        pi(false) = []
        pi(s) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- 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)

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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          p#_A(x1) = x1 + 1
          s_A(x1) = x1
    
      precedence:
      
        p# = s
      
      partial status:
    
        pi(p#) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          p#_A(x1) = x1 + 3
          s_A(x1) = x1 + 1
    
      precedence:
      
        p# > s
      
      partial status:
    
        pi(p#) = []
        pi(s) = [1]
    

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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          intlist#_A(x1) = max{5, x1 + 3}
          if_intlist#_A(x1,x2) = max{5, x1, x2 + 2}
          empty_A(x1) = x1 + 3
          false_A = 6
          tail_A(x1) = max{1, x1 - 2}
          nil_A = 0
          true_A = 2
          cons_A(x1,x2) = max{x1 + 7, x2 + 7}
    
      precedence:
      
        intlist# > if_intlist# = empty = false = tail = nil = true = cons
      
      partial status:
    
        pi(intlist#) = [1]
        pi(if_intlist#) = [1, 2]
        pi(empty) = [1]
        pi(false) = []
        pi(tail) = []
        pi(nil) = []
        pi(true) = []
        pi(cons) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          intlist#_A(x1) = max{7, x1 + 5}
          if_intlist#_A(x1,x2) = max{x1 + 2, x2 + 4}
          empty_A(x1) = max{4, x1 + 2}
          false_A = 5
          tail_A(x1) = 2
          nil_A = 4
          true_A = 5
          cons_A(x1,x2) = max{x1 + 6, x2 + 6}
    
      precedence:
      
        intlist# = if_intlist# = empty = false = tail = nil = true = cons
      
      partial status:
    
        pi(intlist#) = []
        pi(if_intlist#) = [2]
        pi(empty) = []
        pi(false) = []
        pi(tail) = []
        pi(nil) = []
        pi(true) = []
        pi(cons) = [2]
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.