YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: plus#(x,s(y)) -> plus#(x,y)
p2: times#(s(x),y) -> plus#(times(x,y),y)
p3: times#(s(x),y) -> times#(x,y)
p4: p#(s(s(x))) -> p#(s(x))
p5: fac#(s(x)) -> times#(fac(p(s(x))),s(x))
p6: fac#(s(x)) -> fac#(p(s(x)))
p7: fac#(s(x)) -> p#(s(x))

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(x,s(y)) -> s(plus(x,y))
r3: times(|0|(),y) -> |0|()
r4: times(x,|0|()) -> |0|()
r5: times(s(x),y) -> plus(times(x,y),y)
r6: p(s(s(x))) -> s(p(s(x)))
r7: p(s(|0|())) -> |0|()
r8: fac(s(x)) -> times(fac(p(s(x))),s(x))

The estimated dependency graph contains the following SCCs:

  {p6}
  {p3}
  {p1}
  {p4}


-- Reduction pair.

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

p1: fac#(s(x)) -> fac#(p(s(x)))

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(x,s(y)) -> s(plus(x,y))
r3: times(|0|(),y) -> |0|()
r4: times(x,|0|()) -> |0|()
r5: times(s(x),y) -> plus(times(x,y),y)
r6: p(s(s(x))) -> s(p(s(x)))
r7: p(s(|0|())) -> |0|()
r8: fac(s(x)) -> times(fac(p(s(x))),s(x))

The set of usable rules consists of

  r6, r7

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          fac#_A(x1) = max{5, x1 + 3}
          s_A(x1) = x1 + 13
          p_A(x1) = max{8, x1 - 5}
          |0|_A = 7
    
      precedence:
      
        fac# = p > |0| > s
      
      partial status:
    
        pi(fac#) = [1]
        pi(s) = []
        pi(p) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          fac#_A(x1) = 0
          s_A(x1) = 6
          p_A(x1) = 11
          |0|_A = 6
    
      precedence:
      
        s > |0| > fac# = p
      
      partial status:
    
        pi(fac#) = []
        pi(s) = []
        pi(p) = []
        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: times#(s(x),y) -> times#(x,y)

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(x,s(y)) -> s(plus(x,y))
r3: times(|0|(),y) -> |0|()
r4: times(x,|0|()) -> |0|()
r5: times(s(x),y) -> plus(times(x,y),y)
r6: p(s(s(x))) -> s(p(s(x)))
r7: p(s(|0|())) -> |0|()
r8: fac(s(x)) -> times(fac(p(s(x))),s(x))

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

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(x,s(y)) -> s(plus(x,y))
r3: times(|0|(),y) -> |0|()
r4: times(x,|0|()) -> |0|()
r5: times(s(x),y) -> plus(times(x,y),y)
r6: p(s(s(x))) -> s(p(s(x)))
r7: p(s(|0|())) -> |0|()
r8: fac(s(x)) -> times(fac(p(s(x))),s(x))

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.

-- Reduction pair.

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

p1: p#(s(s(x))) -> p#(s(x))

and R consists of:

r1: plus(x,|0|()) -> x
r2: plus(x,s(y)) -> s(plus(x,y))
r3: times(|0|(),y) -> |0|()
r4: times(x,|0|()) -> |0|()
r5: times(s(x),y) -> plus(times(x,y),y)
r6: p(s(s(x))) -> s(p(s(x)))
r7: p(s(|0|())) -> |0|()
r8: fac(s(x)) -> times(fac(p(s(x))),s(x))

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:
        
          p#_A(x1) = x1 + 1
          s_A(x1) = x1
    
      precedence:
      
        p# = s
      
      partial status:
    
        pi(p#) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          p#_A(x1) = x1 + 3
          s_A(x1) = x1 + 1
    
      precedence:
      
        p# > s
      
      partial status:
    
        pi(p#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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