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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = max{0, x1 - 1, x2 - 1}
            s_A(x1) = x1
            minus#!_A(x1,x2) = max{10, x1 + 1}
            s!_A(x1) = 9
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              minus#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 3
              minus#!_A(x1,x2) = x1 + x2 + 4
              s!_A(x1) = 3
        
  
    partial status:
  
      pi(minus#) = [2]
      pi(s) = [1]

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: 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 reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            double#_A(x1) = max{0, x1 - 1}
            s_A(x1) = max{3, x1}
            double#!_A(x1) = max{1, x1 - 4}
            s!_A(x1) = x1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              double#_A(x1) = x1
              s_A(x1) = x1 + 1
              double#!_A(x1) = x1 + 4
              s!_A(x1) = 3
        
  
    partial status:
  
      pi(double#) = []
      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: 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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = 0
            s_A(x1) = max{105, x1 + 97}
            if_A(x1,x2,x3) = max{63, x2 + 32, x3}
            gt_A(x1,x2) = max{95, x1 - 204, x2 - 99}
            not_A(x1) = max{95, x1 - 291}
            id_A(x1) = max{64, x1 + 31}
            true_A = 95
            false_A = 32
            zero_A = 98
            plus#!_A(x1,x2) = max{x1 + 23, x2 + 13}
            s!_A(x1) = x1 + 225
            if!_A(x1,x2,x3) = max{221, x3 - 94}
            gt!_A(x1,x2) = 224
            not!_A(x1) = x1 + 430
            id!_A(x1) = 119
            true!_A = 223
            false!_A = 221
            zero!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = 0
              s_A(x1) = 2
              if_A(x1,x2,x3) = x2 + x3 + 1
              gt_A(x1,x2) = 1
              not_A(x1) = 3
              id_A(x1) = x1 + 1
              true_A() = 1
              false_A() = 1
              zero_A() = 1
              plus#!_A(x1,x2) = 5
              s!_A(x1) = 4
              if!_A(x1,x2,x3) = 2
              gt!_A(x1,x2) = 2
              not!_A(x1) = 0
              id!_A(x1) = 4
              true!_A() = 5
              false!_A() = 1
              zero!_A() = 0
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = []
      pi(if) = [2, 3]
      pi(gt) = []
      pi(not) = [1]
      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: 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

  r11

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 63}
            id_A(x1) = x1 + 62
            s_A(x1) = x1 + 79
            if_A(x1,x2,x3) = max{194, x2 - 10, x3 + 25}
            gt_A(x1,x2) = max{x1 - 39, x2 + 28}
            false_A = 0
            plus#!_A(x1,x2) = max{45, x1 - 63}
            id!_A(x1) = max{44, x1 + 17}
            s!_A(x1) = x1 + 44
            if!_A(x1,x2,x3) = max{x1 + 2, x3 + 8}
            gt!_A(x1,x2) = 44
            false!_A = 85
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = x1
              id_A(x1) = x1 + 3
              s_A(x1) = 3
              if_A(x1,x2,x3) = x3 + 1
              gt_A(x1,x2) = 1
              false_A() = 2
              plus#!_A(x1,x2) = x1 + 2
              id!_A(x1) = 3
              s!_A(x1) = x1 + 4
              if!_A(x1,x2,x3) = 2
              gt!_A(x1,x2) = 4
              false!_A() = 0
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(id) = [1]
      pi(s) = [1]
      pi(if) = [1, 3]
      pi(gt) = [1]
      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: 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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            gt#_A(x1,x2) = max{0, x1 - 5, x2 - 7}
            s_A(x1) = max{9, x1 + 6}
            gt#!_A(x1,x2) = max{1, x1 - 284, x2 - 10}
            s!_A(x1) = max{0, x1 - 285}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              gt#_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 2
              gt#!_A(x1,x2) = x1 + x2 + 2
              s!_A(x1) = 1
        
  
    partial status:
  
      pi(gt#) = [2]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.