YES
We show the termination of the TRS R:

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


-- SCC decomposition.

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

p1: plus#(s(x),y) -> plus#(p(s(x)),y)
p2: plus#(s(x),y) -> p#(s(x))
p3: times#(s(x),y) -> plus#(y,times(p(s(x)),y))
p4: times#(s(x),y) -> times#(p(s(x)),y)
p5: times#(s(x),y) -> p#(s(x))
p6: p#(s(s(x))) -> p#(s(x))
p7: fac#(s(x),y) -> fac#(p(s(x)),times(s(x),y))
p8: fac#(s(x),y) -> p#(s(x))
p9: fac#(s(x),y) -> times#(s(x),y)
p10: factorial#(x) -> fac#(x,s(|0|()))

and R consists of:

r1: plus(|0|(),x) -> x
r2: plus(s(x),y) -> s(plus(p(s(x)),y))
r3: times(|0|(),y) -> |0|()
r4: times(s(x),y) -> plus(y,times(p(s(x)),y))
r5: p(s(|0|())) -> |0|()
r6: p(s(s(x))) -> s(p(s(x)))
r7: fac(|0|(),x) -> x
r8: fac(s(x),y) -> fac(p(s(x)),times(s(x),y))
r9: factorial(x) -> fac(x,s(|0|()))

The estimated dependency graph contains the following SCCs:

  {p7}
  {p4}
  {p1}
  {p6}


-- Reduction pair.

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

p1: fac#(s(x),y) -> fac#(p(s(x)),times(s(x),y))

and R consists of:

r1: plus(|0|(),x) -> x
r2: plus(s(x),y) -> s(plus(p(s(x)),y))
r3: times(|0|(),y) -> |0|()
r4: times(s(x),y) -> plus(y,times(p(s(x)),y))
r5: p(s(|0|())) -> |0|()
r6: p(s(s(x))) -> s(p(s(x)))
r7: fac(|0|(),x) -> x
r8: fac(s(x),y) -> fac(p(s(x)),times(s(x),y))
r9: factorial(x) -> fac(x,s(|0|()))

The set of usable rules consists of

  r5, r6

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,x2) = x1
              s_A(x1) = x1
              p_A(x1) = x1
              times_A(x1,x2) = x2
              |0|_A() = 1
              fac#!_A(x1,x2) = 0
              s!_A(x1) = 3
              p!_A(x1) = 3
              times!_A(x1,x2) = 1
              |0|!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            fac#_A(x1,x2) = max{0, x1 - 26}
            s_A(x1) = x1 + 25
            p_A(x1) = max{1, x1 - 23}
            times_A(x1,x2) = max{72, x1 + 9, x2 - 34}
            |0|_A = 2
            fac#!_A(x1,x2) = max{13, x1 - 3}
            s!_A(x1) = x1 + 37
            p!_A(x1) = max{1, x1 - 4}
            times!_A(x1,x2) = max{x1 + 74, x2 + 68}
            |0|!_A = 69
        
  
    partial status:
  
      pi(fac#) = []
      pi(s) = [1]
      pi(p) = []
      pi(times) = [1]
      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#(p(s(x)),y)

and R consists of:

r1: plus(|0|(),x) -> x
r2: plus(s(x),y) -> s(plus(p(s(x)),y))
r3: times(|0|(),y) -> |0|()
r4: times(s(x),y) -> plus(y,times(p(s(x)),y))
r5: p(s(|0|())) -> |0|()
r6: p(s(s(x))) -> s(p(s(x)))
r7: fac(|0|(),x) -> x
r8: fac(s(x),y) -> fac(p(s(x)),times(s(x),y))
r9: factorial(x) -> fac(x,s(|0|()))

The set of usable rules consists of

  r5, r6

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 + 5
              p_A(x1) = x1
              |0|_A() = 2
              times#!_A(x1,x2) = 5
              s!_A(x1) = 4
              p!_A(x1) = 6
              |0|!_A() = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times#_A(x1,x2) = max{0, x1 - 26, x2 - 28}
            s_A(x1) = x1 + 25
            p_A(x1) = max{2, x1 - 23}
            |0|_A = 12
            times#!_A(x1,x2) = max{19, x1 - 5}
            s!_A(x1) = max{12, x1 + 6}
            p!_A(x1) = max{4, x1 - 4}
            |0|!_A = 14
        
  
    partial status:
  
      pi(times#) = []
      pi(s) = [1]
      pi(p) = [1]
      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: plus#(s(x),y) -> plus#(p(s(x)),y)

and R consists of:

r1: plus(|0|(),x) -> x
r2: plus(s(x),y) -> s(plus(p(s(x)),y))
r3: times(|0|(),y) -> |0|()
r4: times(s(x),y) -> plus(y,times(p(s(x)),y))
r5: p(s(|0|())) -> |0|()
r6: p(s(s(x))) -> s(p(s(x)))
r7: fac(|0|(),x) -> x
r8: fac(s(x),y) -> fac(p(s(x)),times(s(x),y))
r9: factorial(x) -> fac(x,s(|0|()))

The set of usable rules consists of

  r5, r6

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) = x1
              s_A(x1) = 3
              p_A(x1) = 3
              |0|_A() = 0
              plus#!_A(x1,x2) = x1 + 1
              s!_A(x1) = 3
              p!_A(x1) = x1
              |0|!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = max{20, x1 + 4, x2}
            s_A(x1) = x1 + 29
            p_A(x1) = max{16, x1 - 13}
            |0|_A = 17
            plus#!_A(x1,x2) = max{99, x1 + 4}
            s!_A(x1) = x1 + 98
            p!_A(x1) = x1 + 56
            |0|!_A = 65
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = []
      pi(p) = []
      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: p#(s(s(x))) -> p#(s(x))

and R consists of:

r1: plus(|0|(),x) -> x
r2: plus(s(x),y) -> s(plus(p(s(x)),y))
r3: times(|0|(),y) -> |0|()
r4: times(s(x),y) -> plus(y,times(p(s(x)),y))
r5: p(s(|0|())) -> |0|()
r6: p(s(s(x))) -> s(p(s(x)))
r7: fac(|0|(),x) -> x
r8: fac(s(x),y) -> fac(p(s(x)),times(s(x),y))
r9: factorial(x) -> fac(x,s(|0|()))

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) = 3
              p#!_A(x1) = 2
              s!_A(x1) = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            p#_A(x1) = x1
            s_A(x1) = x1 + 24
            p#!_A(x1) = max{41, x1 - 7}
            s!_A(x1) = max{9, x1 - 5}
        
  
    partial status:
  
      pi(p#) = [1]
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.