YES

We show the termination of the TRS R:

  plus(plus(X,Y),Z) -> plus(X,plus(Y,Z))
  times(X,s(Y)) -> plus(X,times(Y,X))

-- SCC decomposition.

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

p1: plus#(plus(X,Y),Z) -> plus#(X,plus(Y,Z))
p2: plus#(plus(X,Y),Z) -> plus#(Y,Z)
p3: times#(X,s(Y)) -> plus#(X,times(Y,X))
p4: times#(X,s(Y)) -> times#(Y,X)

and R consists of:

r1: plus(plus(X,Y),Z) -> plus(X,plus(Y,Z))
r2: times(X,s(Y)) -> plus(X,times(Y,X))

The estimated dependency graph contains the following SCCs:

  {p4}
  {p1, p2}


-- Reduction pair.

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

p1: times#(X,s(Y)) -> times#(Y,X)

and R consists of:

r1: plus(plus(X,Y),Z) -> plus(X,plus(Y,Z))
r2: times(X,s(Y)) -> plus(X,times(Y,X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      times#_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2
      s_A(x1) = ((1,1),(1,1)) x1 + (1,0)

The next rules are strictly ordered:

  p1
  r1, r2

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: plus#(plus(X,Y),Z) -> plus#(X,plus(Y,Z))
p2: plus#(plus(X,Y),Z) -> plus#(Y,Z)

and R consists of:

r1: plus(plus(X,Y),Z) -> plus(X,plus(Y,Z))
r2: times(X,s(Y)) -> plus(X,times(Y,X))

The set of usable rules consists of

  r1

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      plus#_A(x1,x2) = ((0,1),(1,1)) x1 + ((0,0),(1,1)) x2
      plus_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,0),(1,1)) x2 + (1,1)

The next rules are strictly ordered:

  p1, p2

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