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:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = max{2, x1}
        s_A(x1) = max{9, x1}
        plus_A(x1,x2) = max{11, x1 + 4, x2 + 3}
        minus_A(x1,x2) = max{9, x1}
        double_A(x1) = max{7, x1}
        |0|_A = 8
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = 0
          s_A(x1) = max{5, x1}
          plus_A(x1,x2) = max{7, x2 + 2}
          minus_A(x1,x2) = x1 + 7
          double_A(x1) = max{4, x1}
          |0|_A = 5
          plus#!_A(x1,x2) = 1
          s!_A(x1) = 2
          plus!_A(x1,x2) = 4
          minus!_A(x1,x2) = 0
          double!_A(x1) = 3
          |0|!_A = 2
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]
      pi(plus) = [1]
      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:
  
      max/plus interpretation on natural numbers:
      
        minus#_A(x1,x2) = max{1, x1 - 2, x2 - 8}
        s_A(x1) = x1 + 7
  
    base order 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 - 9}
          minus#!_A(x1,x2) = max{2, x1 - 5, x2 - 21}
          s!_A(x1) = max{1, x1 - 20}
  
    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)))
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:
  
      max/plus interpretation on natural numbers:
      
        double#_A(x1) = 0
        s_A(x1) = max{11, x1 - 14}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          double#_A(x1) = 0
          s_A(x1) = max{11, x1 + 5}
          double#!_A(x1) = max{3, x1 - 1}
          s!_A(x1) = max{2, x1 - 49}
  
    partial status:
  
      pi(double#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.