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 - 25, x2 - 25}
            s_A(x1) = x1 + 24
            minus#!_A(x1,x2) = max{2, x1 - 21, x2 - 161}
            s!_A(x1) = max{1, x1 - 162}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = max{0, x1 - 6, x2 - 6}
            s_A(x1) = max{32, x1 + 5}
            minus#!_A(x1,x2) = max{4, x1 - 33}
            s!_A(x1) = max{3, x1 - 165}
        
  
    partial status:
  
      pi(minus#) = []
      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: 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 - 23}
            s_A(x1) = max{42, x1 + 22}
            double#!_A(x1) = max{1, x1 - 40}
            s!_A(x1) = x1 + 121
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            double#_A(x1) = max{0, x1 - 8}
            s_A(x1) = x1 + 7
            double#!_A(x1) = max{1, x1 - 8}
            s!_A(x1) = 0
        
  
    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) = max{x1, x2 - 8}
            s_A(x1) = x1 + 9
            if_A(x1,x2,x3) = max{x1, x2, x3}
            gt_A(x1,x2) = max{x1 - 26, x2}
            not_A(x1) = max{7, x1}
            id_A(x1) = x1
            true_A = 0
            false_A = 0
            zero_A = 0
            plus#!_A(x1,x2) = max{0, x1 - 4, x2 - 12}
            s!_A(x1) = max{27, x1 + 23}
            if!_A(x1,x2,x3) = max{x1 + 24, x3 + 36}
            gt!_A(x1,x2) = max{28, x1 - 13, x2 - 13}
            not!_A(x1) = max{37, x1 + 25}
            id!_A(x1) = max{11, x1 - 15}
            true!_A = 26
            false!_A = 26
            zero!_A = 27
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 92, x2 - 102}
            s_A(x1) = x1 + 41
            if_A(x1,x2,x3) = max{90, x1 + 11, x2 + 1, x3}
            gt_A(x1,x2) = max{21, x1 - 42, x2 - 72}
            not_A(x1) = max{90, x1 + 20}
            id_A(x1) = max{79, x1 + 35}
            true_A = 21
            false_A = 21
            zero_A = 0
            plus#!_A(x1,x2) = max{49, x1 - 446, x2 - 447}
            s!_A(x1) = max{48, x1 - 97}
            if!_A(x1,x2,x3) = max{51, x1 - 332, x2 - 483, x3 - 496}
            gt!_A(x1,x2) = max{51, x1 - 96, x2 - 322}
            not!_A(x1) = max{52, x1 - 24}
            id!_A(x1) = max{50, x1 - 96}
            true!_A = 0
            false!_A = 0
            zero!_A = 1
        
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(if) = [2, 3]
      pi(gt) = [2]
      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

  r10, r11, 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) = max{0, x1 - 40, x2 - 22}
            id_A(x1) = x1
            s_A(x1) = max{14, x1 + 12}
            if_A(x1,x2,x3) = max{16, x1 - 1, x2 + 12, x3}
            gt_A(x1,x2) = max{22, x1 - 30, x2 - 18}
            true_A = 14
            false_A = 2
            zero_A = 0
            plus#!_A(x1,x2) = max{39, x1 - 72, x2 - 54}
            id!_A(x1) = max{40, x1 - 73}
            s!_A(x1) = 40
            if!_A(x1,x2,x3) = max{29, x1 - 23, x2 - 3, x3 - 15}
            gt!_A(x1,x2) = max{29, x1 - 88, x2 - 15}
            true!_A = 1
            false!_A = 0
            zero!_A = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 91, x2 - 60}
            id_A(x1) = max{28, x1 + 15}
            s_A(x1) = max{26, x1 + 16}
            if_A(x1,x2,x3) = max{33, x1 - 1, x2 + 16, x3}
            gt_A(x1,x2) = max{60, x1 - 27, x2 - 17}
            true_A = 60
            false_A = 60
            zero_A = 0
            plus#!_A(x1,x2) = max{3, x1 - 117, x2 - 60}
            id!_A(x1) = max{1, x1 - 89}
            s!_A(x1) = max{5, x1 - 121}
            if!_A(x1,x2,x3) = max{2, x1 - 95, x2 - 122, x3 - 114}
            gt!_A(x1,x2) = max{4, x1 - 114, x2 - 27}
            true!_A = 0
            false!_A = 0
            zero!_A = 1
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(id) = [1]
      pi(s) = [1]
      pi(if) = [2, 3]
      pi(gt) = [2]
      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: 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 - 62, x2 - 62}
            s_A(x1) = x1 + 61
            gt#!_A(x1,x2) = max{4, x1 - 62, x2 - 94}
            s!_A(x1) = max{3, x1 - 94}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            gt#_A(x1,x2) = max{0, x1 - 19, x2 - 19}
            s_A(x1) = x1 + 18
            gt#!_A(x1,x2) = max{1, x1 - 16, x2 - 99}
            s!_A(x1) = max{2, x1 - 98}
        
  
    partial status:
  
      pi(gt#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.