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

  r13, r14

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        minus#_A(x1,x2) = max{x1 - 1, x2 + 15}
        s_A(x1) = x1 + 15
        gt_A(x1,x2) = max{5, x1 - 23, x2 - 5}
        zero_A = 4
        true_A = 3
        false_A = 3
  
    base order 2:
  
      precedence:
      
        minus# > s = gt = zero > true = false
      
  
    partial status:
  
      pi(minus#) = [2]
      pi(s) = [1]
      pi(gt) = []
      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: 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

  r13, r14

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        double#_A(x1) = max{8, x1 - 1}
        s_A(x1) = x1 + 43
        gt_A(x1,x2) = max{65, x1 - 44, x2 - 59}
        zero_A = 43
        true_A = 11
        false_A = 0
  
    base order 2:
  
      precedence:
      
        gt = zero > s = false > double# = true
      
  
    partial status:
  
      pi(double#) = []
      pi(s) = [1]
      pi(gt) = []
      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: 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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = max{34, x1 - 42, x2 - 47}
        s_A(x1) = x1 + 82
        if_A(x1,x2,x3) = max{32, x1 - 34, x2 + 21, x3}
        gt_A(x1,x2) = max{30, x1 - 207, x2 - 207}
        not_A(x1) = max{33, x1 - 31}
        id_A(x1) = max{20, x1 + 1}
        true_A = 29
        false_A = 11
        zero_A = 30
  
    base order 2:
  
      precedence:
      
        not > if > plus# = s > gt = zero > id = true = false
      
  
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(if) = [3]
      pi(gt) = []
      pi(not) = []
      pi(id) = [1]
      pi(true) = []
      pi(false) = []
      pi(zero) = []

The following rules are removed:

  p1



-- 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:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = max{8, x1 + 4}
        id_A(x1) = max{9, x1 + 5}
        s_A(x1) = x1 + 63
        if_A(x1,x2,x3) = max{7, x1 - 41, x2 + 1, x3 + 1}
        gt_A(x1,x2) = max{6, x1 - 201, x2 - 138}
        true_A = 0
        false_A = 0
        zero_A = 1
  
    base order 2:
  
      precedence:
      
        if > id = s > plus# = gt = zero > true = false
      
  
    partial status:
  
      pi(plus#) = [1]
      pi(id) = [1]
      pi(s) = [1]
      pi(if) = []
      pi(gt) = []
      pi(true) = []
      pi(false) = []
      pi(zero) = []

The following rules are removed:

  p2



-- 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

  r13, r14

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = x1 + 1
        id_A(x1) = max{4, x1 + 2}
        s_A(x1) = max{22, x1 - 33}
        if_A(x1,x2,x3) = max{11, x1 - 40, x2 - 119, x3 - 86}
        gt_A(x1,x2) = max{6, x1 - 46, x2 - 79}
        zero_A = 6
        true_A = 5
        false_A = 0
  
    base order 2:
  
      precedence:
      
        plus# = id = s > if = gt = zero > true = false
      
  
    partial status:
  
      pi(plus#) = [1]
      pi(id) = []
      pi(s) = []
      pi(if) = []
      pi(gt) = []
      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

  r13, r14

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        gt#_A(x1,x2) = max{x1 - 1, x2 + 62}
        s_A(x1) = max{69, x1 + 62}
        gt_A(x1,x2) = max{68, x1 - 70, x2 - 72}
        zero_A = 64
        true_A = 63
        false_A = 33
  
    base order 2:
  
      precedence:
      
        gt# > s > gt = zero > true = false
      
  
    partial status:
  
      pi(gt#) = [2]
      pi(s) = []
      pi(gt) = []
      pi(zero) = []
      pi(true) = []
      pi(false) = []

The following rules are removed:

  p1

Then no dependency pair remains.