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) = x1
            s_A(x1) = max{7, x1}
            plus_A(x1,x2) = max{1, x1, x2}
            minus_A(x1,x2) = max{3, x1}
            double_A(x1) = max{2, x1}
            |0|_A = 3
            plus#!_A(x1,x2) = 7
            s!_A(x1) = 7
            plus!_A(x1,x2) = max{15, x2 + 8}
            minus!_A(x1,x2) = 1
            double!_A(x1) = max{6, x1 + 1}
            |0|!_A = 5
        
        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) = x2
              s_A(x1) = x1
              plus_A(x1,x2) = x2 + 1
              minus_A(x1,x2) = x1 + x2 + 1
              double_A(x1) = x1
              |0|_A() = 1
              plus#!_A(x1,x2) = 2
              s!_A(x1) = 4
              plus!_A(x1,x2) = 3
              minus!_A(x1,x2) = 1
              double!_A(x1) = 1
              |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) = 0
            s_A(x1) = max{25, x1 + 19}
            minus#!_A(x1,x2) = max{5, x2 - 26}
            s!_A(x1) = max{5, x1 + 1}
        
        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
              minus#!_A(x1,x2) = 2
              s!_A(x1) = 2
        
  
    partial status:
  
      pi(minus#) = [2]
      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) = x1 + 14
            double#!_A(x1) = max{0, x1 - 15}
            s!_A(x1) = max{0, x1 - 3}
        
        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
              double#!_A(x1) = x1 + 2
              s!_A(x1) = x1 + 2
        
  
    partial status:
  
      pi(double#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.