YES
We show the termination of the TRS R:

  times(x,plus(y,|1|())) -> plus(times(x,plus(y,times(|1|(),|0|()))),x)
  times(x,|1|()) -> x
  times(x,|0|()) -> |0|()
  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,plus(y,|1|())) -> plus#(times(x,plus(y,times(|1|(),|0|()))),x)
p2: times#(x,plus(y,|1|())) -> times#(x,plus(y,times(|1|(),|0|())))
p3: times#(x,plus(y,|1|())) -> plus#(y,times(|1|(),|0|()))
p4: times#(x,plus(y,|1|())) -> times#(|1|(),|0|())
p5: plus#(s(x),s(y)) -> plus#(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))
p6: plus#(s(x),s(y)) -> if#(gt(x,y),x,y)
p7: plus#(s(x),s(y)) -> gt#(x,y)
p8: plus#(s(x),s(y)) -> if#(not(gt(x,y)),id(x),id(y))
p9: plus#(s(x),s(y)) -> not#(gt(x,y))
p10: plus#(s(x),s(y)) -> id#(x)
p11: plus#(s(x),s(y)) -> id#(y)
p12: plus#(s(x),x) -> plus#(if(gt(x,x),id(x),id(x)),s(x))
p13: plus#(s(x),x) -> if#(gt(x,x),id(x),id(x))
p14: plus#(s(x),x) -> gt#(x,x)
p15: plus#(s(x),x) -> id#(x)
p16: plus#(id(x),s(y)) -> plus#(x,if(gt(s(y),y),y,s(y)))
p17: plus#(id(x),s(y)) -> if#(gt(s(y),y),y,s(y))
p18: plus#(id(x),s(y)) -> gt#(s(y),y)
p19: not#(x) -> if#(x,false(),true())
p20: gt#(s(x),s(y)) -> gt#(x,y)

and R consists of:

r1: times(x,plus(y,|1|())) -> plus(times(x,plus(y,times(|1|(),|0|()))),x)
r2: times(x,|1|()) -> x
r3: times(x,|0|()) -> |0|()
r4: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r5: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r6: plus(zero(),y) -> y
r7: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r8: id(x) -> x
r9: if(true(),x,y) -> x
r10: if(false(),x,y) -> y
r11: not(x) -> if(x,false(),true())
r12: gt(s(x),zero()) -> true()
r13: gt(zero(),y) -> false()
r14: gt(s(x),s(y)) -> gt(x,y)

The estimated dependency graph contains the following SCCs:

  {p2}
  {p5, p12, p16}
  {p20}


-- Reduction pair.

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

p1: times#(x,plus(y,|1|())) -> times#(x,plus(y,times(|1|(),|0|())))

and R consists of:

r1: times(x,plus(y,|1|())) -> plus(times(x,plus(y,times(|1|(),|0|()))),x)
r2: times(x,|1|()) -> x
r3: times(x,|0|()) -> |0|()
r4: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r5: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r6: plus(zero(),y) -> y
r7: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r8: id(x) -> x
r9: if(true(),x,y) -> x
r10: if(false(),x,y) -> y
r11: not(x) -> if(x,false(),true())
r12: gt(s(x),zero()) -> true()
r13: gt(zero(),y) -> false()
r14: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times#_A(x1,x2) = max{94, x1 + 16, x2 - 40}
        plus_A(x1,x2) = max{17, x1 + 10, x2 + 10}
        |1|_A = 29
        times_A(x1,x2) = max{70, x1 - 92, x2 + 42}
        |0|_A = 51
        s_A(x1) = max{3, x1}
        if_A(x1,x2,x3) = max{7, x1 - 5, x2, x3}
        gt_A(x1,x2) = max{3, x1 - 9, x2 - 4}
        not_A(x1) = x1 + 8
        id_A(x1) = max{4, x1}
        zero_A = 3
        true_A = 2
        false_A = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          times#_A(x1,x2) = max{0, x1 - 93, x2 - 72}
          plus_A(x1,x2) = max{69, x2 + 1}
          |1|_A = 70
          times_A(x1,x2) = max{x1 - 91, x2 + 42}
          |0|_A = 24
          s_A(x1) = 19
          if_A(x1,x2,x3) = max{x1 - 9, x2 + 25, x3 + 22}
          gt_A(x1,x2) = 34
          not_A(x1) = max{34, x1}
          id_A(x1) = x1 + 22
          zero_A = 16
          true_A = 12
          false_A = 9
          times#!_A(x1,x2) = max{2, x2 - 68}
          plus!_A(x1,x2) = 0
          |1|!_A = 0
          times!_A(x1,x2) = max{0, x1 - 74, x2 - 72}
          |0|!_A = 4
          s!_A(x1) = 0
          if!_A(x1,x2,x3) = max{2, x2 - 23, x3 - 23}
          gt!_A(x1,x2) = 0
          not!_A(x1) = max{1, x1 - 15}
          id!_A(x1) = max{0, x1 - 5}
          zero!_A = 1
          true!_A = 0
          false!_A = 0
  
    partial status:
  
      pi(times#) = [1]
      pi(plus) = []
      pi(|1|) = []
      pi(times) = [2]
      pi(|0|) = []
      pi(s) = []
      pi(if) = [2, 3]
      pi(gt) = []
      pi(not) = []
      pi(id) = [1]
      pi(zero) = []
      pi(true) = []
      pi(false) = []

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,plus(y,|1|())) -> plus(times(x,plus(y,times(|1|(),|0|()))),x)
r2: times(x,|1|()) -> x
r3: times(x,|0|()) -> |0|()
r4: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r5: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r6: plus(zero(),y) -> y
r7: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r8: id(x) -> x
r9: if(true(),x,y) -> x
r10: if(false(),x,y) -> y
r11: not(x) -> if(x,false(),true())
r12: gt(s(x),zero()) -> true()
r13: gt(zero(),y) -> false()
r14: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r8, r9, r10, r11, r12, r13, r14

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = max{1, x1 - 204, x2 - 210}
        s_A(x1) = max{28, x1 + 6}
        if_A(x1,x2,x3) = max{66, x1 - 68, x2 + 4, x3}
        gt_A(x1,x2) = max{66, x1 - 39, x2 - 68}
        not_A(x1) = max{67, x1 - 67}
        id_A(x1) = max{50, x1 + 2}
        true_A = 7
        false_A = 0
        zero_A = 38
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = max{0, x1 - 114, x2 - 114}
          s_A(x1) = x1 + 92
          if_A(x1,x2,x3) = max{43, x1 - 39, x2, x3}
          gt_A(x1,x2) = max{82, x1 - 93, x2 - 93}
          not_A(x1) = max{152, x1 + 40}
          id_A(x1) = max{69, x1}
          true_A = 82
          false_A = 82
          zero_A = 0
          plus#!_A(x1,x2) = max{1, x1 - 90, x2 - 114}
          s!_A(x1) = max{2, x1 - 1818}
          if!_A(x1,x2,x3) = max{3, x1 - 173, x2 - 184, x3 - 2396}
          gt!_A(x1,x2) = max{2, x1 - 2528, x2 - 2395}
          not!_A(x1) = max{4, x1 - 103}
          id!_A(x1) = max{2, x1 - 2459}
          true!_A = 0
          false!_A = 0
          zero!_A = 1
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(if) = [3]
      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,plus(y,|1|())) -> plus(times(x,plus(y,times(|1|(),|0|()))),x)
r2: times(x,|1|()) -> x
r3: times(x,|0|()) -> |0|()
r4: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r5: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r6: plus(zero(),y) -> y
r7: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r8: id(x) -> x
r9: if(true(),x,y) -> x
r10: if(false(),x,y) -> y
r11: not(x) -> if(x,false(),true())
r12: gt(s(x),zero()) -> true()
r13: gt(zero(),y) -> false()
r14: 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,plus(y,|1|())) -> plus(times(x,plus(y,times(|1|(),|0|()))),x)
r2: times(x,|1|()) -> x
r3: times(x,|0|()) -> |0|()
r4: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r5: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r6: plus(zero(),y) -> y
r7: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r8: id(x) -> x
r9: if(true(),x,y) -> x
r10: if(false(),x,y) -> y
r11: not(x) -> if(x,false(),true())
r12: gt(s(x),zero()) -> true()
r13: gt(zero(),y) -> false()
r14: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r9, r10, r12, r13, r14

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = x1 + 2
        id_A(x1) = max{2, x1}
        s_A(x1) = x1
        if_A(x1,x2,x3) = max{x2 + 7, x3}
        gt_A(x1,x2) = max{5, x1 - 3, x2 - 3}
        true_A = 1
        false_A = 0
        zero_A = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = max{0, x1 - 268, x2 - 65}
          id_A(x1) = max{21, x1 + 18}
          s_A(x1) = max{20, x1 + 18}
          if_A(x1,x2,x3) = max{19, x1 - 1, x2 + 18, x3}
          gt_A(x1,x2) = max{65, x1 - 200, x2 - 19}
          true_A = 65
          false_A = 20
          zero_A = 0
          plus#!_A(x1,x2) = max{46, x1 - 197, x2 - 239}
          id!_A(x1) = max{58, x1 - 421}
          s!_A(x1) = max{57, x1 + 19}
          if!_A(x1,x2,x3) = max{18, x1 - 18, x2 - 200, x3 - 200}
          gt!_A(x1,x2) = max{0, x1 - 220, x2 - 220}
          true!_A = 58
          false!_A = 0
          zero!_A = 59
  
    partial status:
  
      pi(plus#) = [1]
      pi(id) = [1]
      pi(s) = []
      pi(if) = [2, 3]
      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,plus(y,|1|())) -> plus(times(x,plus(y,times(|1|(),|0|()))),x)
r2: times(x,|1|()) -> x
r3: times(x,|0|()) -> |0|()
r4: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r5: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r6: plus(zero(),y) -> y
r7: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r8: id(x) -> x
r9: if(true(),x,y) -> x
r10: if(false(),x,y) -> y
r11: not(x) -> if(x,false(),true())
r12: gt(s(x),zero()) -> true()
r13: gt(zero(),y) -> false()
r14: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r9, r10, r12, r13, r14

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        gt#_A(x1,x2) = max{57, x1 - 57, x2 - 57}
        s_A(x1) = max{56, x1 + 9}
        if_A(x1,x2,x3) = max{141, x1 - 22, x2 + 16, x3 + 10}
        true_A = 5
        false_A = 5
        gt_A(x1,x2) = max{6, x1 - 72, x2 - 143}
        zero_A = 71
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          gt#_A(x1,x2) = max{0, x1 - 31, x2 - 31}
          s_A(x1) = x1 + 30
          if_A(x1,x2,x3) = max{41, x1 - 1, x2 + 10, x3 + 7}
          true_A = 31
          false_A = 31
          gt_A(x1,x2) = max{41, x1 - 55, x2 - 31}
          zero_A = 0
          gt#!_A(x1,x2) = max{3, x1 - 693, x2 - 26}
          s!_A(x1) = max{2, x1 - 693}
          if!_A(x1,x2,x3) = 2
          true!_A = 0
          false!_A = 0
          gt!_A(x1,x2) = max{3, x1 - 562, x2 - 563}
          zero!_A = 1
  
    partial status:
  
      pi(gt#) = []
      pi(s) = []
      pi(if) = []
      pi(true) = []
      pi(false) = []
      pi(gt) = []
      pi(zero) = []

The following rules are removed:

  p1

Then no dependency pair remains.