YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: bin#(s(x),s(y)) -> bin#(x,s(y))
p2: bin#(s(x),s(y)) -> bin#(x,y)

and R consists of:

r1: bin(x,|0|()) -> s(|0|())
r2: bin(|0|(),s(y)) -> |0|()
r3: bin(s(x),s(y)) -> +(bin(x,s(y)),bin(x,y))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: bin#(s(x),s(y)) -> bin#(x,s(y))
p2: bin#(s(x),s(y)) -> bin#(x,y)

and R consists of:

r1: bin(x,|0|()) -> s(|0|())
r2: bin(|0|(),s(y)) -> |0|()
r3: bin(s(x),s(y)) -> +(bin(x,s(y)),bin(x,y))

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

The next rules are strictly ordered:

  p1, p2

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