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:
    
        max/plus interpretations on natural numbers:
        
          times#_A(x1,x2) = max{1, x2 - 6}
          s_A(x1) = max{5, x1}
          plus_A(x1,x2) = max{14, x1 + 8, x2 + 4}
          times_A(x1,x2) = max{x1, x2 + 10}
          |0|_A = 0
    
      precedence:
      
        times# = plus > s > times = |0|
      
      partial status:
    
        pi(times#) = []
        pi(s) = [1]
        pi(plus) = [2]
        pi(times) = [1, 2]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          times#_A(x1,x2) = 0
          s_A(x1) = x1 + 15
          plus_A(x1,x2) = max{5, x2 - 4}
          times_A(x1,x2) = max{x1 + 15, x2 + 2}
          |0|_A = 14
    
      precedence:
      
        s = times = |0| > plus > times#
      
      partial status:
    
        pi(times#) = []
        pi(s) = [1]
        pi(plus) = []
        pi(times) = [1, 2]
        pi(|0|) = []
    

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: times#(x,s(y)) -> times#(x,y)
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:

  {p1, p2}


-- 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,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:
    
        max/plus interpretations on natural numbers:
        
          times#_A(x1,x2) = max{x1 + 13, x2 + 4}
          s_A(x1) = max{6, x1}
          plus_A(x1,x2) = max{x1 + 14, x2 + 7}
          times_A(x1,x2) = max{x1, x2 + 6}
          |0|_A = 0
    
      precedence:
      
        times# = plus > s > times = |0|
      
      partial status:
    
        pi(times#) = [2]
        pi(s) = [1]
        pi(plus) = [2]
        pi(times) = [2]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          times#_A(x1,x2) = x2 + 3
          s_A(x1) = max{12, x1 + 8}
          plus_A(x1,x2) = max{5, x2}
          times_A(x1,x2) = max{6, x2}
          |0|_A = 7
    
      precedence:
      
        times# = s = plus = times = |0|
      
      partial status:
    
        pi(times#) = []
        pi(s) = []
        pi(plus) = [2]
        pi(times) = [2]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2

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

The next rules are strictly ordered:

  p1

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