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:
  
      max/plus interpretation on natural numbers:
      
        fac#_A(x1,x2) = max{1, x1 - 17}
        s_A(x1) = max{39, x1 + 16}
        p_A(x1) = max{21, x1 - 7}
        times_A(x1,x2) = max{19, x1 + 2}
        |0|_A = 40
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          fac#_A(x1,x2) = max{0, x1 - 9}
          s_A(x1) = max{8, x1 - 2}
          p_A(x1) = max{1, x1}
          times_A(x1,x2) = 0
          |0|_A = 8
          fac#!_A(x1,x2) = 27
          s!_A(x1) = max{24, x1 - 14}
          p!_A(x1) = max{25, x1 - 8}
          times!_A(x1,x2) = max{26, x1 + 3, x2 + 13}
          |0|!_A = 14
  
    partial status:
  
      pi(fac#) = []
      pi(s) = []
      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:
  
      max/plus interpretation on natural numbers:
      
        times#_A(x1,x2) = max{x1 + 33, x2 + 12}
        s_A(x1) = max{10, x1 - 21}
        p_A(x1) = max{10, x1 - 14}
        |0|_A = 9
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          times#_A(x1,x2) = max{0, x1 - 50, x2 - 54}
          s_A(x1) = x1 + 49
          p_A(x1) = max{25, x1 - 24}
          |0|_A = 16
          times#!_A(x1,x2) = max{30, x1 - 50, x2 - 54}
          s!_A(x1) = max{29, x1 + 2}
          p!_A(x1) = max{28, x1 - 68}
          |0|!_A = 19
  
    partial status:
  
      pi(times#) = [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: 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:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = 0
        s_A(x1) = max{10, x1 - 18}
        p_A(x1) = max{14, x1 - 29}
        |0|_A = 9
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = max{0, x1 - 15, x2 - 17}
          s_A(x1) = x1 + 14
          p_A(x1) = max{1, x1 - 13}
          |0|_A = 16
          plus#!_A(x1,x2) = max{9, x1 - 2}
          s!_A(x1) = max{7, x1 - 34}
          p!_A(x1) = max{8, x1 - 31}
          |0|!_A = 0
  
    partial status:
  
      pi(plus#) = []
      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:
  
      max/plus interpretation on natural numbers:
      
        p#_A(x1) = max{1, x1 - 9}
        s_A(x1) = max{11, x1 + 8}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          p#_A(x1) = 0
          s_A(x1) = max{44, x1 - 13}
          p#!_A(x1) = 2
          s!_A(x1) = max{3, x1 - 43}
  
    partial status:
  
      pi(p#) = []
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.