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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        times#_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(1,1)) x2
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        times#_A(x1,x2) = (0,0)
        s_A(x1) = (1,1)
    

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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,1),(0,1)) x2
        plus_A(x1,x2) = x1 + ((0,1),(0,1)) x2 + (2,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = (0,0)
        plus_A(x1,x2) = (1,1)
    

The next rules are strictly ordered:

  p2

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#(X,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#(X,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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((0,1),(0,0)) x1
        plus_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,1),(0,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = (0,0)
        plus_A(x1,x2) = (1,1)
    

The next rules are strictly ordered:

  p1

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