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:

  max/plus interpretations on natural numbers:
  
    times#_A(x1,x2) = max{0, x2 - 2}
    plus_A(x1,x2) = max{3, x2}
    |1|_A = 4
    times_A(x1,x2) = max{x1 - 3, x2}
    |0|_A = 1
    id_A(x1) = x1
    if_A(x1,x2,x3) = max{x1 - 1, x2, x3}
    true_A = 0
    false_A = 0
    not_A(x1) = max{0, x1 - 1}
    gt_A(x1,x2) = 0
    s_A(x1) = 1
    zero_A = 0

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: 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:

  max/plus interpretations on natural numbers:
  
    plus#_A(x1,x2) = max{x1 + 6, x2 + 6}
    s_A(x1) = x1 + 1
    if_A(x1,x2,x3) = max{x2, x3}
    gt_A(x1,x2) = max{x1, x2 + 1}
    not_A(x1) = max{2, x1 - 3}
    id_A(x1) = x1
    true_A = 1
    false_A = 1
    zero_A = 1

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- 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)))
p2: 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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- 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)))
p2: 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, r12, r13, r14

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    plus#_A(x1,x2) = max{x1 + 11, x2 + 7}
    id_A(x1) = max{3, x1}
    s_A(x1) = x1 + 4
    if_A(x1,x2,x3) = max{3, x1 - 6, x2, x3}
    gt_A(x1,x2) = max{x1 - 2, x2 + 1}
    true_A = 2
    false_A = 1
    zero_A = 1

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- 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:

  max/plus interpretations on natural numbers:
  
    plus#_A(x1,x2) = max{2, x1}
    id_A(x1) = x1 + 5
    s_A(x1) = x1 + 6
    if_A(x1,x2,x3) = max{x1, x2, x3 + 1}
    gt_A(x1,x2) = max{x1 - 4, x2 + 5}
    zero_A = 7
    false_A = 3
    true_A = 0

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: 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

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    gt#_A(x1,x2) = max{x1 + 1, x2 + 1}
    s_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

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