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 path order with precedence:
  
    precedence:
    
      bin# > s
    
    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.