YES

We show the termination of the TRS R:

  times(x,plus(y,s(z))) -> plus(times(x,plus(y,times(s(z),|0|()))),times(x,s(z)))
  times(x,|0|()) -> |0|()
  times(x,s(y)) -> plus(times(x,y),x)
  plus(x,|0|()) -> x
  plus(x,s(y)) -> s(plus(x,y))

-- SCC decomposition.

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

p1: times#(x,plus(y,s(z))) -> plus#(times(x,plus(y,times(s(z),|0|()))),times(x,s(z)))
p2: times#(x,plus(y,s(z))) -> times#(x,plus(y,times(s(z),|0|())))
p3: times#(x,plus(y,s(z))) -> plus#(y,times(s(z),|0|()))
p4: times#(x,plus(y,s(z))) -> times#(s(z),|0|())
p5: times#(x,plus(y,s(z))) -> times#(x,s(z))
p6: times#(x,s(y)) -> plus#(times(x,y),x)
p7: times#(x,s(y)) -> times#(x,y)
p8: plus#(x,s(y)) -> plus#(x,y)

and R consists of:

r1: times(x,plus(y,s(z))) -> plus(times(x,plus(y,times(s(z),|0|()))),times(x,s(z)))
r2: times(x,|0|()) -> |0|()
r3: times(x,s(y)) -> plus(times(x,y),x)
r4: plus(x,|0|()) -> x
r5: plus(x,s(y)) -> s(plus(x,y))

The estimated dependency graph contains the following SCCs:

  {p2, p5, p7}
  {p8}


-- Reduction pair.

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

p1: times#(x,s(y)) -> times#(x,y)
p2: times#(x,plus(y,s(z))) -> times#(x,s(z))
p3: times#(x,plus(y,s(z))) -> times#(x,plus(y,times(s(z),|0|())))

and R consists of:

r1: times(x,plus(y,s(z))) -> plus(times(x,plus(y,times(s(z),|0|()))),times(x,s(z)))
r2: times(x,|0|()) -> |0|()
r3: times(x,s(y)) -> plus(times(x,y),x)
r4: plus(x,|0|()) -> x
r5: plus(x,s(y)) -> s(plus(x,y))

The set of usable rules consists of

  r2, r4, r5

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            times#_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (6,1)
            s_A(x1) = x1 + (2,0)
            plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (3,2)
            times_A(x1,x2) = (2,1)
            |0|_A() = (1,0)
    
      precedence:
      
        times# = s = plus = times = |0|
      
      partial status:
    
        pi(times#) = []
        pi(s) = []
        pi(plus) = []
        pi(times) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            times#_A(x1,x2) = x1 + (3,5)
            s_A(x1) = ((1,0),(1,0)) x1 + (0,1)
            plus_A(x1,x2) = ((1,0),(1,1)) x1 + (2,4)
            times_A(x1,x2) = (0,0)
            |0|_A() = (1,2)
    
      precedence:
      
        times > |0| > times# > s = plus
      
      partial status:
    
        pi(times#) = [1]
        pi(s) = []
        pi(plus) = [1]
        pi(times) = []
        pi(|0|) = []
    

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: times#(x,plus(y,s(z))) -> times#(x,s(z))
p2: times#(x,plus(y,s(z))) -> times#(x,plus(y,times(s(z),|0|())))

and R consists of:

r1: times(x,plus(y,s(z))) -> plus(times(x,plus(y,times(s(z),|0|()))),times(x,s(z)))
r2: times(x,|0|()) -> |0|()
r3: times(x,s(y)) -> plus(times(x,y),x)
r4: plus(x,|0|()) -> x
r5: plus(x,s(y)) -> s(plus(x,y))

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: times#(x,plus(y,s(z))) -> times#(x,plus(y,times(s(z),|0|())))

and R consists of:

r1: times(x,plus(y,s(z))) -> plus(times(x,plus(y,times(s(z),|0|()))),times(x,s(z)))
r2: times(x,|0|()) -> |0|()
r3: times(x,s(y)) -> plus(times(x,y),x)
r4: plus(x,|0|()) -> x
r5: plus(x,s(y)) -> s(plus(x,y))

The set of usable rules consists of

  r2, r4, r5

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            times#_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,2)
            plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (1,4)
            s_A(x1) = x1 + (5,3)
            times_A(x1,x2) = (2,1)
            |0|_A() = (1,1)
    
      precedence:
      
        plus > times# = s = times = |0|
      
      partial status:
    
        pi(times#) = []
        pi(plus) = [1]
        pi(s) = [1]
        pi(times) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            times#_A(x1,x2) = (0,0)
            plus_A(x1,x2) = ((1,0),(1,1)) x1 + (3,4)
            s_A(x1) = ((1,0),(1,1)) x1 + (2,3)
            times_A(x1,x2) = (0,1)
            |0|_A() = (1,2)
    
      precedence:
      
        plus > s = times > times# = |0|
      
      partial status:
    
        pi(times#) = []
        pi(plus) = [1]
        pi(s) = [1]
        pi(times) = []
        pi(|0|) = []
    

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#(x,s(y)) -> plus#(x,y)

and R consists of:

r1: times(x,plus(y,s(z))) -> plus(times(x,plus(y,times(s(z),|0|()))),times(x,s(z)))
r2: times(x,|0|()) -> |0|()
r3: times(x,s(y)) -> plus(times(x,y),x)
r4: plus(x,|0|()) -> x
r5: plus(x,s(y)) -> s(plus(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            plus#_A(x1,x2) = ((0,0),(1,0)) x2 + (1,1)
            s_A(x1) = x1 + (2,2)
    
      precedence:
      
        s > plus#
      
      partial status:
    
        pi(plus#) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            plus#_A(x1,x2) = (0,0)
            s_A(x1) = (1,1)
    
      precedence:
      
        plus# = s
      
      partial status:
    
        pi(plus#) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

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