YES

We show the termination of the TRS R:

  minus(x,|0|()) -> x
  minus(s(x),s(y)) -> minus(x,y)
  double(|0|()) -> |0|()
  double(s(x)) -> s(s(double(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: minus#(s(x),s(y)) -> minus#(x,y)
p2: double#(s(x)) -> double#(x)
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: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: double(|0|()) -> |0|()
r4: double(s(x)) -> s(s(double(x)))
r5: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r7: plus(zero(),y) -> y
r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r9: id(x) -> x
r10: if(true(),x,y) -> x
r11: if(false(),x,y) -> y
r12: not(x) -> if(x,false(),true())
r13: gt(s(x),zero()) -> true()
r14: gt(zero(),y) -> false()
r15: gt(s(x),s(y)) -> gt(x,y)

The estimated dependency graph contains the following SCCs:

  {p1}
  {p2}
  {p3, p10, p14}
  {p18}


-- Reduction pair.

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

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

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: double(|0|()) -> |0|()
r4: double(s(x)) -> s(s(double(x)))
r5: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r7: plus(zero(),y) -> y
r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r9: id(x) -> x
r10: if(true(),x,y) -> x
r11: if(false(),x,y) -> y
r12: not(x) -> if(x,false(),true())
r13: gt(s(x),zero()) -> true()
r14: gt(zero(),y) -> false()
r15: 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:
  
    minus#_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.

-- Reduction pair.

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

p1: double#(s(x)) -> double#(x)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: double(|0|()) -> |0|()
r4: double(s(x)) -> s(s(double(x)))
r5: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r7: plus(zero(),y) -> y
r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r9: id(x) -> x
r10: if(true(),x,y) -> x
r11: if(false(),x,y) -> y
r12: not(x) -> if(x,false(),true())
r13: gt(s(x),zero()) -> true()
r14: gt(zero(),y) -> false()
r15: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  max/plus interpretations on natural numbers:
  
    double#_A(x1) = x1
    s_A(x1) = x1 + 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: 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: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: double(|0|()) -> |0|()
r4: double(s(x)) -> s(s(double(x)))
r5: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r7: plus(zero(),y) -> y
r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r9: id(x) -> x
r10: if(true(),x,y) -> x
r11: if(false(),x,y) -> y
r12: not(x) -> if(x,false(),true())
r13: gt(s(x),zero()) -> true()
r14: gt(zero(),y) -> false()
r15: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r9, r10, r11, r12, r13, r14, r15

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: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: double(|0|()) -> |0|()
r4: double(s(x)) -> s(s(double(x)))
r5: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r7: plus(zero(),y) -> y
r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r9: id(x) -> x
r10: if(true(),x,y) -> x
r11: if(false(),x,y) -> y
r12: not(x) -> if(x,false(),true())
r13: gt(s(x),zero()) -> true()
r14: gt(zero(),y) -> false()
r15: 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: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: double(|0|()) -> |0|()
r4: double(s(x)) -> s(s(double(x)))
r5: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r7: plus(zero(),y) -> y
r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r9: id(x) -> x
r10: if(true(),x,y) -> x
r11: if(false(),x,y) -> y
r12: not(x) -> if(x,false(),true())
r13: gt(s(x),zero()) -> true()
r14: gt(zero(),y) -> false()
r15: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r9, r10, r11, r13, r14, r15

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: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: double(|0|()) -> |0|()
r4: double(s(x)) -> s(s(double(x)))
r5: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r7: plus(zero(),y) -> y
r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r9: id(x) -> x
r10: if(true(),x,y) -> x
r11: if(false(),x,y) -> y
r12: not(x) -> if(x,false(),true())
r13: gt(s(x),zero()) -> true()
r14: gt(zero(),y) -> false()
r15: 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: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: double(|0|()) -> |0|()
r4: double(s(x)) -> s(s(double(x)))
r5: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r7: plus(zero(),y) -> y
r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r9: id(x) -> x
r10: if(true(),x,y) -> x
r11: if(false(),x,y) -> y
r12: not(x) -> if(x,false(),true())
r13: gt(s(x),zero()) -> true()
r14: gt(zero(),y) -> false()
r15: gt(s(x),s(y)) -> gt(x,y)

The set of usable rules consists of

  r10, r11, r13, r14, r15

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: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: double(|0|()) -> |0|()
r4: double(s(x)) -> s(s(double(x)))
r5: plus(s(x),s(y)) -> s(s(plus(if(gt(x,y),x,y),if(not(gt(x,y)),id(x),id(y)))))
r6: plus(s(x),x) -> plus(if(gt(x,x),id(x),id(x)),s(x))
r7: plus(zero(),y) -> y
r8: plus(id(x),s(y)) -> s(plus(x,if(gt(s(y),y),y,s(y))))
r9: id(x) -> x
r10: if(true(),x,y) -> x
r11: if(false(),x,y) -> y
r12: not(x) -> if(x,false(),true())
r13: gt(s(x),zero()) -> true()
r14: gt(zero(),y) -> false()
r15: 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.