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(|0|(),y) -> y
  plus(s(x),y) -> s(plus(x,y))
  plus(s(x),y) -> plus(x,s(y))
  plus(s(x),y) -> s(plus(minus(x,y),double(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),y) -> plus#(x,y)
p4: plus#(s(x),y) -> plus#(x,s(y))
p5: plus#(s(x),y) -> plus#(minus(x,y),double(y))
p6: plus#(s(x),y) -> minus#(x,y)
p7: plus#(s(x),y) -> double#(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(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: plus(s(x),y) -> plus(x,s(y))
r8: plus(s(x),y) -> s(plus(minus(x,y),double(y)))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p5}
  {p1}
  {p2}


-- Reduction pair.

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

p1: plus#(s(x),y) -> plus#(minus(x,y),double(y))
p2: plus#(s(x),y) -> plus#(x,s(y))
p3: plus#(s(x),y) -> plus#(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(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: plus(s(x),y) -> plus(x,s(y))
r8: plus(s(x),y) -> s(plus(minus(x,y),double(y)))

The set of usable rules consists of

  r1, r2, r3, r4

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 - 9, x2 - 5}
            s_A(x1) = max{4, x1}
            minus_A(x1,x2) = max{5, x1, x2 + 4}
            double_A(x1) = max{4, x1 - 5}
            |0|_A = 0
            plus#!_A(x1,x2) = 0
            s!_A(x1) = max{2, x1 - 12}
            minus!_A(x1,x2) = max{1, x1 - 13}
            double!_A(x1) = max{2, x1 - 5}
            |0|!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{0, x1 - 9, x2 - 4}
            s_A(x1) = max{3, x1}
            minus_A(x1,x2) = max{6, x1, x2 - 8}
            double_A(x1) = max{3, x1 - 4}
            |0|_A = 3
            plus#!_A(x1,x2) = max{x1 + 7, x2 + 2}
            s!_A(x1) = max{6, x1 - 8}
            minus!_A(x1,x2) = max{5, x1 - 16}
            double!_A(x1) = max{11, x1 - 4}
            |0|!_A = 0
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(minus) = [1]
      pi(double) = [1]
      pi(|0|) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.

-- 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(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: plus(s(x),y) -> plus(x,s(y))
r8: plus(s(x),y) -> s(plus(minus(x,y),double(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) = 0
            s_A(x1) = max{25, x1 + 7}
            minus#!_A(x1,x2) = max{1, x2 - 5}
            s!_A(x1) = max{22, x1 - 4}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = 0
            s_A(x1) = max{4, x1 - 2}
            minus#!_A(x1,x2) = max{7, x2 + 3}
            s!_A(x1) = max{8, x1 + 6}
        
  
    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(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: plus(s(x),y) -> plus(x,s(y))
r8: plus(s(x),y) -> s(plus(minus(x,y),double(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) = 0
            s_A(x1) = max{2, x1 - 3}
            double#!_A(x1) = max{1, x1 - 7}
            s!_A(x1) = max{1, x1 - 6}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            double#_A(x1) = 0
            s_A(x1) = max{2, x1 - 3}
            double#!_A(x1) = max{1, x1 - 5}
            s!_A(x1) = max{1, x1 - 5}
        
  
    partial status:
  
      pi(double#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.