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)))
  plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z))


-- 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)
p8: plus#(s(plus(x,y)),z) -> plus#(plus(x,y),z)

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)))
r9: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p5, p8}
  {p1}
  {p2}


-- Reduction pair.

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

p1: plus#(s(plus(x,y)),z) -> plus#(plus(x,y),z)
p2: plus#(s(x),y) -> plus#(minus(x,y),double(y))
p3: plus#(s(x),y) -> plus#(x,s(y))
p4: 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)))
r9: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9

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, x2 - 31}
            s_A(x1) = max{19, x1}
            plus_A(x1,x2) = max{x1 + 25, x2 + 23}
            minus_A(x1,x2) = max{19, x1}
            double_A(x1) = max{13, x1}
            |0|_A = 14
            plus#!_A(x1,x2) = max{20, x1 + 1}
            s!_A(x1) = max{51, x1 + 2}
            plus!_A(x1,x2) = max{56, x1 + 30, x2 + 37}
            minus!_A(x1,x2) = max{50, x1 - 20}
            double!_A(x1) = max{56, x1 + 37}
            |0|!_A = 52
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{x1 - 1, x2 + 6}
            s_A(x1) = x1
            plus_A(x1,x2) = max{8, x1 - 9, x2 + 6}
            minus_A(x1,x2) = max{2, x1, x2 - 15}
            double_A(x1) = max{0, x1 - 20}
            |0|_A = 0
            plus#!_A(x1,x2) = max{x1 + 14, x2 + 17}
            s!_A(x1) = max{18, x1 - 12}
            plus!_A(x1,x2) = max{x1 - 5, x2 + 19}
            minus!_A(x1,x2) = max{20, x1 - 6, x2 - 1}
            double!_A(x1) = max{3, x1 - 11}
            |0|!_A = 0
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(plus) = [1, 2]
      pi(minus) = [1]
      pi(double) = [1]
      pi(|0|) = []

The following rules are removed:

  p1, p2, p3, p4

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)))
r9: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z))

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{x1, x2}
            s_A(x1) = x1 + 31
            minus#!_A(x1,x2) = max{3, x1 - 32, x2 - 27}
            s!_A(x1) = max{2, x1 - 9}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = 0
            s_A(x1) = max{3, x1 - 31}
            minus#!_A(x1,x2) = max{1, x1 - 1, x2 - 23}
            s!_A(x1) = max{7, x1 - 23}
        
  
    partial status:
  
      pi(minus#) = [1]
      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(|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)))
r9: plus(s(plus(x,y)),z) -> s(plus(plus(x,y),z))

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 - 15}
            double#!_A(x1) = 1
            s!_A(x1) = max{3, x1 - 40}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            double#_A(x1) = 0
            s_A(x1) = max{4, x1 - 9}
            double#!_A(x1) = 0
            s!_A(x1) = max{2, x1 - 6}
        
  
    partial status:
  
      pi(double#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.