YES

We show the termination of the TRS R:

  double(|0|()) -> |0|()
  double(s(x)) -> s(s(double(x)))
  +(x,|0|()) -> x
  +(x,s(y)) -> s(+(x,y))
  +(s(x),y) -> s(+(x,y))
  double(x) -> +(x,x)

-- SCC decomposition.

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

p1: double#(s(x)) -> double#(x)
p2: +#(x,s(y)) -> +#(x,y)
p3: +#(s(x),y) -> +#(x,y)
p4: double#(x) -> +#(x,x)

and R consists of:

r1: double(|0|()) -> |0|()
r2: double(s(x)) -> s(s(double(x)))
r3: +(x,|0|()) -> x
r4: +(x,s(y)) -> s(+(x,y))
r5: +(s(x),y) -> s(+(x,y))
r6: double(x) -> +(x,x)

The estimated dependency graph contains the following SCCs:

  {p1}
  {p2, p3}


-- Reduction pair.

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

p1: double#(s(x)) -> double#(x)

and R consists of:

r1: double(|0|()) -> |0|()
r2: double(s(x)) -> s(s(double(x)))
r3: +(x,|0|()) -> x
r4: +(x,s(y)) -> s(+(x,y))
r5: +(s(x),y) -> s(+(x,y))
r6: double(x) -> +(x,x)

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

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: +#(x,s(y)) -> +#(x,y)
p2: +#(s(x),y) -> +#(x,y)

and R consists of:

r1: double(|0|()) -> |0|()
r2: double(s(x)) -> s(s(double(x)))
r3: +(x,|0|()) -> x
r4: +(x,s(y)) -> s(+(x,y))
r5: +(s(x),y) -> s(+(x,y))
r6: double(x) -> +(x,x)

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 > +#
      
      argument filter:
    
        pi(+#) = [1, 2]
        pi(s) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > +#
      
      argument filter:
    
        pi(+#) = 2
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: +#(s(x),y) -> +#(x,y)

and R consists of:

r1: double(|0|()) -> |0|()
r2: double(s(x)) -> s(s(double(x)))
r3: +(x,|0|()) -> x
r4: +(x,s(y)) -> s(+(x,y))
r5: +(s(x),y) -> s(+(x,y))
r6: double(x) -> +(x,x)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: +#(s(x),y) -> +#(x,y)

and R consists of:

r1: double(|0|()) -> |0|()
r2: double(s(x)) -> s(s(double(x)))
r3: +(x,|0|()) -> x
r4: +(x,s(y)) -> s(+(x,y))
r5: +(s(x),y) -> s(+(x,y))
r6: double(x) -> +(x,x)

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 > +#
      
      argument filter:
    
        pi(+#) = 1
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > +#
      
      argument filter:
    
        pi(+#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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