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

  r4, r6, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              fac#_A(x1) = x1
              s_A(x1) = x1 + 3
              p_A(x1) = x1
              times_A(x1,x2) = x1 + x2 + 1
              |0|_A() = 4
              fac#!_A(x1) = x1
              s!_A(x1) = x1 + 1
              p!_A(x1) = 2
              times!_A(x1,x2) = x1 + x2
              |0|!_A() = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            fac#_A(x1) = x1 + 8
            s_A(x1) = x1 + 28
            p_A(x1) = max{2, x1 - 19}
            times_A(x1,x2) = max{x1 - 37, x2 + 10}
            |0|_A = 11
            fac#!_A(x1) = max{1, x1 - 10}
            s!_A(x1) = max{22, x1 + 6}
            p!_A(x1) = max{5, x1 - 33}
            times!_A(x1,x2) = max{24, x1 - 66, x2 + 9}
            |0|!_A = 21
        
  
    partial status:
  
      pi(fac#) = [1]
      pi(s) = [1]
      pi(p) = [1]
      pi(times) = [1, 2]
      pi(|0|) = []

The following rules are removed:

  p1

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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              times#_A(x1,x2) = x1
              s_A(x1) = x1 + 1
              times#!_A(x1,x2) = 0
              s!_A(x1) = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times#_A(x1,x2) = 0
            s_A(x1) = max{18, x1 + 7}
            times#!_A(x1,x2) = max{1, x1 - 5}
            s!_A(x1) = max{5, x1 - 5}
        
  
    partial status:
  
      pi(times#) = []
      pi(s) = []

The following rules are removed:

  p1

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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = x2
              s_A(x1) = x1 + 1
              plus#!_A(x1,x2) = x2 + 1
              s!_A(x1) = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = 0
            s_A(x1) = max{7, x1 - 18}
            plus#!_A(x1,x2) = max{1, x1 - 1, x2 - 62}
            s!_A(x1) = max{3, x1 - 63}
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = []

The following rules are removed:

  p1

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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              p#_A(x1) = x1
              s_A(x1) = x1 + 1
              p#!_A(x1) = 3
              s!_A(x1) = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            p#_A(x1) = 0
            s_A(x1) = max{2, x1 - 9}
            p#!_A(x1) = max{3, x1 - 25}
            s!_A(x1) = max{3, x1 - 24}
        
  
    partial status:
  
      pi(p#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.