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 reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        times#_A(x1,x2) = max{x1 + 1, x2 - 1}
        s_A(x1) = x1 + 3
  
    precedence:
    
      times# = s
    
    partial status:
  
      pi(times#) = []
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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 monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        plus#_A(x1,x2) = max{x1 + 4, x2 + 2}
        plus_A(x1,x2) = max{x1 + 2, x2}
  
    precedence:
    
      plus# = plus
    
    partial status:
  
      pi(plus#) = [1, 2]
      pi(plus) = [1, 2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        plus#_A(x1,x2) = max{x1 + 1, x2 + 1}
        plus_A(x1,x2) = x2
  
    precedence:
    
      plus# = plus
    
    partial status:
  
      pi(plus#) = [1]
      pi(plus) = [2]

The next rules are strictly ordered:

  p1

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