YES

We show the termination of the TRS R:

  plus(s(X),plus(Y,Z)) -> plus(X,plus(s(s(Y)),Z))
  plus(s(X1),plus(X2,plus(X3,X4))) -> plus(X1,plus(X3,plus(X2,X4)))

-- SCC decomposition.

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

p1: plus#(s(X),plus(Y,Z)) -> plus#(X,plus(s(s(Y)),Z))
p2: plus#(s(X),plus(Y,Z)) -> plus#(s(s(Y)),Z)
p3: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X1,plus(X3,plus(X2,X4)))
p4: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X3,plus(X2,X4))
p5: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X2,X4)

and R consists of:

r1: plus(s(X),plus(Y,Z)) -> plus(X,plus(s(s(Y)),Z))
r2: plus(s(X1),plus(X2,plus(X3,X4))) -> plus(X1,plus(X3,plus(X2,X4)))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: plus#(s(X),plus(Y,Z)) -> plus#(X,plus(s(s(Y)),Z))
p2: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X2,X4)
p3: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X3,plus(X2,X4))
p4: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X1,plus(X3,plus(X2,X4)))
p5: plus#(s(X),plus(Y,Z)) -> plus#(s(s(Y)),Z)

and R consists of:

r1: plus(s(X),plus(Y,Z)) -> plus(X,plus(s(s(Y)),Z))
r2: plus(s(X1),plus(X2,plus(X3,X4))) -> plus(X1,plus(X3,plus(X2,X4)))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

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

The next rules are strictly ordered:

  p2, p3, p5

We remove them from the problem.

-- SCC decomposition.

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

p1: plus#(s(X),plus(Y,Z)) -> plus#(X,plus(s(s(Y)),Z))
p2: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X1,plus(X3,plus(X2,X4)))

and R consists of:

r1: plus(s(X),plus(Y,Z)) -> plus(X,plus(s(s(Y)),Z))
r2: plus(s(X1),plus(X2,plus(X3,X4))) -> plus(X1,plus(X3,plus(X2,X4)))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: plus#(s(X),plus(Y,Z)) -> plus#(X,plus(s(s(Y)),Z))
p2: plus#(s(X1),plus(X2,plus(X3,X4))) -> plus#(X1,plus(X3,plus(X2,X4)))

and R consists of:

r1: plus(s(X),plus(Y,Z)) -> plus(X,plus(s(s(Y)),Z))
r2: plus(s(X1),plus(X2,plus(X3,X4))) -> plus(X1,plus(X3,plus(X2,X4)))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

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

The next rules are strictly ordered:

  p1, p2

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