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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        |0| > s > double > minus > plus > plus#
      
      argument filter:
    
        pi(plus#) = [1, 2]
        pi(s) = 1
        pi(plus) = [1, 2]
        pi(minus) = 1
        pi(double) = 1
        pi(|0|) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |0| > double > minus > plus > s > plus#
      
      argument filter:
    
        pi(plus#) = 1
        pi(s) = [1]
        pi(plus) = [1]
        pi(minus) = 1
        pi(double) = [1]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4

We remove them from the problem.  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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > minus#
      
      argument filter:
    
        pi(minus#) = [2]
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > minus#
      
      argument filter:
    
        pi(minus#) = [2]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  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 monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > double#
      
      argument filter:
    
        pi(double#) = [1]
        pi(s) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > double#
      
      argument filter:
    
        pi(double#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9

We remove them from the problem.  Then no dependency pair remains.