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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      plus#_A(x1,x2) = ((0,0,0),(1,0,0),(0,1,0)) x2
      s_A(x1) = ((0,0,0),(0,0,0),(1,0,0)) x1 + (1,0,1)
      plus_A(x1,x2) = ((1,0,0),(1,1,0),(1,0,0)) x2 + (1,1,1)

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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      plus#_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,0)) x1 + ((0,0,0),(1,0,0),(0,1,0)) x2
      s_A(x1) = ((1,0,0),(1,1,0),(0,1,1)) x1 + (1,3,1)
      plus_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,0)) x1 + ((0,0,0),(1,0,0),(0,0,0)) x2 + (1,1,1)

The next rules are strictly ordered:

  p1, p2

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