YES
We show the termination of the TRS R:

  times(x,|0|()) -> |0|()
  times(x,s(y)) -> plus(times(x,y),x)
  plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
  plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
  plus(zero(),y) -> y
  plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
  id(x) -> x
  if(true(),x,y) -> x
  if(false(),x,y) -> y
  not(x) -> if(x,false(),true())
  gt(s(x),zero()) -> true()
  gt(zero(),y) -> false()
  gt(s(x),s(y)) -> gt(x,y)


-- SCC decomposition.

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

p1: times#(x,s(y)) -> plus#(times(x,y),x)
p2: times#(x,s(y)) -> times#(x,y)
p3: plus#(s(x),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p4: plus#(s(x),s(y)) -> if#(gt(x,y),x,y)
p5: plus#(s(x),s(y)) -> gt#(x,y)
p6: plus#(s(x),s(y)) -> if#(not(gt(x,y)),id(x),id(y))
p7: plus#(s(x),s(y)) -> not#(gt(x,y))
p8: plus#(s(x),s(y)) -> id#(x)
p9: plus#(s(x),s(y)) -> id#(y)
p10: plus#(s(x),x) -> plus#(if(gt(x,x),id(x),id(x)),s(x))
p11: plus#(s(x),x) -> if#(gt(x,x),id(x),id(x))
p12: plus#(s(x),x) -> gt#(x,x)
p13: plus#(s(x),x) -> id#(x)
p14: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))
p15: plus#(id(x),s(y)) -> if#(gt(s(y),y),y,s(y))
p16: plus#(id(x),s(y)) -> gt#(s(y),y)
p17: not#(x) -> if#(x,false(),true())
p18: gt#(s(x),s(y)) -> gt#(x,y)

and R consists of:

r1: times(x,|0|()) -> |0|()
r2: times(x,s(y)) -> plus(times(x,y),x)
r3: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r4: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r5: plus(zero(),y) -> y
r6: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r7: id(x) -> x
r8: if(true(),x,y) -> x
r9: if(false(),x,y) -> y
r10: not(x) -> if(x,false(),true())
r11: gt(s(x),zero()) -> true()
r12: gt(zero(),y) -> false()
r13: gt(s(x),s(y)) -> gt(x,y)

The estimated dependency graph contains the following SCCs:

  {p2}
  {p3, p10, p14}
  {p18}


-- Reduction pair.

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

p1: times#(x,s(y)) -> times#(x,y)

and R consists of:

r1: times(x,|0|()) -> |0|()
r2: times(x,s(y)) -> plus(times(x,y),x)
r3: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r4: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r5: plus(zero(),y) -> y
r6: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r7: id(x) -> x
r8: if(true(),x,y) -> x
r9: if(false(),x,y) -> y
r10: not(x) -> if(x,false(),true())
r11: gt(s(x),zero()) -> true()
r12: gt(zero(),y) -> false()
r13: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times#_A(x1,x2) = max{0, x1 - 1}
        s_A(x1) = max{0, x1 - 2}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          times#_A(x1,x2) = max{0, x2 - 4}
          s_A(x1) = x1 + 3
          times#!_A(x1,x2) = max{1, x2 - 1}
          s!_A(x1) = 0
  
    partial status:
  
      pi(times#) = []
      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: plus#(s(x),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p2: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))
p3: plus#(s(x),x) -> plus#(if(gt(x,x),id(x),id(x)),s(x))

and R consists of:

r1: times(x,|0|()) -> |0|()
r2: times(x,s(y)) -> plus(times(x,y),x)
r3: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r4: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r5: plus(zero(),y) -> y
r6: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r7: id(x) -> x
r8: if(true(),x,y) -> x
r9: if(false(),x,y) -> y
r10: not(x) -> if(x,false(),true())
r11: gt(s(x),zero()) -> true()
r12: gt(zero(),y) -> false()
r13: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r7, r8, r9, r10, r11, r12, r13

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = 0
        s_A(x1) = x1 + 31
        if_A(x1,x2,x3) = max{23, x1 - 13, x2 + 11, x3 + 5}
        gt_A(x1,x2) = max{24, x1 - 7, x2 - 32}
        not_A(x1) = max{24, x1 - 12}
        id_A(x1) = max{22, x1}
        true_A = 5
        false_A = 0
        zero_A = 6
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = max{0, x1 - 49, x2 - 53}
          s_A(x1) = max{47, x1 + 27}
          if_A(x1,x2,x3) = max{44, x1 - 1, x2 + 8, x3}
          gt_A(x1,x2) = max{44, x1 - 48, x2 - 50}
          not_A(x1) = max{53, x1 - 1}
          id_A(x1) = max{32, x1 + 16}
          true_A = 44
          false_A = 0
          zero_A = 0
          plus#!_A(x1,x2) = max{1, x1 - 45, x2 - 71}
          s!_A(x1) = max{4, x1 - 311}
          if!_A(x1,x2,x3) = max{5, x1 - 98, x2 - 33, x3 - 102}
          gt!_A(x1,x2) = max{5, x1 - 92, x2 - 72}
          not!_A(x1) = max{6, x1 - 61}
          id!_A(x1) = max{3, x1 - 17}
          true!_A = 0
          false!_A = 0
          zero!_A = 1
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(if) = []
      pi(gt) = []
      pi(not) = []
      pi(id) = [1]
      pi(true) = []
      pi(false) = []
      pi(zero) = []

The following rules are removed:

  p1, p3



-- SCC decomposition.

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

p1: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))

and R consists of:

r1: times(x,|0|()) -> |0|()
r2: times(x,s(y)) -> plus(times(x,y),x)
r3: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r4: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r5: plus(zero(),y) -> y
r6: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r7: id(x) -> x
r8: if(true(),x,y) -> x
r9: if(false(),x,y) -> y
r10: not(x) -> if(x,false(),true())
r11: gt(s(x),zero()) -> true()
r12: gt(zero(),y) -> false()
r13: gt(s(x),s(y)) -> gt(x,y)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))

and R consists of:

r1: times(x,|0|()) -> |0|()
r2: times(x,s(y)) -> plus(times(x,y),x)
r3: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r4: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r5: plus(zero(),y) -> y
r6: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r7: id(x) -> x
r8: if(true(),x,y) -> x
r9: if(false(),x,y) -> y
r10: not(x) -> if(x,false(),true())
r11: gt(s(x),zero()) -> true()
r12: gt(zero(),y) -> false()
r13: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r8, r9, r11, r12, r13

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = 0
        id_A(x1) = max{1, x1 - 19}
        s_A(x1) = x1 + 17
        if_A(x1,x2,x3) = max{89, x1 - 91, x2 + 18, x3 + 8}
        gt_A(x1,x2) = max{16, x1 - 51, x2 - 142}
        true_A = 15
        false_A = 0
        zero_A = 50
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = max{0, x1 - 63, x2 - 29}
          id_A(x1) = max{9, x1 + 7}
          s_A(x1) = x1 + 28
          if_A(x1,x2,x3) = max{8, x1 - 12, x2, x3}
          gt_A(x1,x2) = max{19, x1 - 37, x2 - 37}
          true_A = 19
          false_A = 8
          zero_A = 0
          plus#!_A(x1,x2) = max{1, x1 - 6}
          id!_A(x1) = max{9, x1 - 84}
          s!_A(x1) = x1 + 7
          if!_A(x1,x2,x3) = max{10, x1 - 59, x2 - 10, x3 - 47}
          gt!_A(x1,x2) = max{6, x1 - 39, x2 - 40}
          true!_A = 5
          false!_A = 0
          zero!_A = 6
  
    partial status:
  
      pi(plus#) = []
      pi(id) = []
      pi(s) = [1]
      pi(if) = [2]
      pi(gt) = []
      pi(true) = []
      pi(false) = []
      pi(zero) = []

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: gt#(s(x),s(y)) -> gt#(x,y)

and R consists of:

r1: times(x,|0|()) -> |0|()
r2: times(x,s(y)) -> plus(times(x,y),x)
r3: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r4: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r5: plus(zero(),y) -> y
r6: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r7: id(x) -> x
r8: if(true(),x,y) -> x
r9: if(false(),x,y) -> y
r10: not(x) -> if(x,false(),true())
r11: gt(s(x),zero()) -> true()
r12: gt(zero(),y) -> false()
r13: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        gt#_A(x1,x2) = max{1, x1 - 4, x2 - 1}
        s_A(x1) = x1 + 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          gt#_A(x1,x2) = max{0, x1 - 4, x2 - 4}
          s_A(x1) = x1 + 3
          gt#!_A(x1,x2) = max{3, x1 - 4, x2 - 4}
          s!_A(x1) = 2
  
    partial status:
  
      pi(gt#) = []
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.