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

  r7

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        minus#_A(x1,x2) = max{242, x1 - 242, x2 - 242}
        s_A(x1) = max{241, x1 + 20}
        plus_A(x1,x2) = max{x1 + 130, x2 + 147}
        zero_A = 228
  
    base order 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 + 1
            plus_A(x1,x2) = x2
            zero_A() = 1
            minus#!_A(x1,x2) = x1 + x2
            s!_A(x1) = x1 + 3
            plus!_A(x1,x2) = 0
            zero!_A() = 1
  
    partial status:
  
      pi(minus#) = []
      pi(s) = [1]
      pi(plus) = [2]
      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: 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

  r7

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        double#_A(x1) = max{1, x1 - 1}
        s_A(x1) = x1
        plus_A(x1,x2) = max{x1 + 22, x2 + 11}
        zero_A = 4
  
    base order 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
            plus_A(x1,x2) = x2
            zero_A() = 1
            double#!_A(x1) = x1 + 2
            s!_A(x1) = x1 + 1
            plus!_A(x1,x2) = 2
            zero!_A() = 3
  
    partial status:
  
      pi(double#) = []
      pi(s) = [1]
      pi(plus) = [2]
      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: 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

  r7, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = max{x1 + 2, x2 - 12}
        s_A(x1) = x1 + 47
        if_A(x1,x2,x3) = max{x2 + 11, x3}
        gt_A(x1,x2) = max{47, x1 - 80, x2 - 47}
        not_A(x1) = 34
        id_A(x1) = max{11, x1}
        plus_A(x1,x2) = max{3, x2 + 2}
        zero_A = 78
        true_A = 32
        false_A = 1
  
    base order 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) = x1 + 1
            if_A(x1,x2,x3) = x2 + x3 + 1
            gt_A(x1,x2) = 1
            not_A(x1) = 3
            id_A(x1) = x1 + 1
            plus_A(x1,x2) = x1 + x2
            zero_A() = 1
            true_A() = 1
            false_A() = 1
            plus#!_A(x1,x2) = 2
            s!_A(x1) = x1 + 4
            if!_A(x1,x2,x3) = 2
            gt!_A(x1,x2) = x2 + 1
            not!_A(x1) = 3
            id!_A(x1) = 5
            plus!_A(x1,x2) = 0
            zero!_A() = 1
            true!_A() = 0
            false!_A() = 0
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(if) = [2, 3]
      pi(gt) = []
      pi(not) = []
      pi(id) = [1]
      pi(plus) = [2]
      pi(zero) = []
      pi(true) = []
      pi(false) = []

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

  r7, r10, r11

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) = 0
        s_A(x1) = max{6, x1 - 30}
        if_A(x1,x2,x3) = max{70, x1 - 3, x2 + 18, x3 + 18}
        gt_A(x1,x2) = max{70, x1 - 2, x2 - 9}
        plus_A(x1,x2) = max{39, x1 - 11, x2 + 1}
        zero_A = 1
        true_A = 0
        false_A = 0
  
    base order 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 + 1
            s_A(x1) = x1 + 1
            if_A(x1,x2,x3) = x2 + x3 + 1
            gt_A(x1,x2) = x1 + 1
            plus_A(x1,x2) = x2
            zero_A() = 1
            true_A() = 3
            false_A() = 3
            plus#!_A(x1,x2) = x1
            id!_A(x1) = x1 + 2
            s!_A(x1) = 1
            if!_A(x1,x2,x3) = 4
            gt!_A(x1,x2) = x1 + x2 + 4
            plus!_A(x1,x2) = 3
            zero!_A() = 1
            true!_A() = 0
            false!_A() = 0
  
    partial status:
  
      pi(plus#) = []
      pi(id) = []
      pi(s) = []
      pi(if) = [2, 3]
      pi(gt) = []
      pi(plus) = [2]
      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: 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:
  
      max/plus interpretation on natural numbers:
      
        gt#_A(x1,x2) = max{12, x1 + 11, x2 - 12}
        s_A(x1) = x1 + 11
  
    base order 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 + 1
            gt#!_A(x1,x2) = x1 + x2 + 4
            s!_A(x1) = 3
  
    partial status:
  
      pi(gt#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.