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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        fac#_A(x1) = max{7, x1 + 2}
        s_A(x1) = max{21, x1 + 11}
        p_A(x1) = max{4, x1 - 11}
        times_A(x1,x2) = max{2, x1 - 22, x2 - 22}
        |0|_A = 1
  
    base order 2:
  
      precedence:
      
        fac# = p > s > times > |0|
      
  
    partial status:
  
      pi(fac#) = [1]
      pi(s) = []
      pi(p) = []
      pi(times) = []
      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

  r4

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times#_A(x1,x2) = max{x1 + 7, x2 + 8}
        s_A(x1) = max{28, x1 + 6}
        times_A(x1,x2) = max{2, x1 - 2, x2 - 29}
        |0|_A = 1
  
    base order 2:
  
      precedence:
      
        times# > s > times > |0|
      
  
    partial status:
  
      pi(times#) = [1]
      pi(s) = [1]
      pi(times) = []
      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#(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

  r4

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus#_A(x1,x2) = max{x1 + 9, x2 + 4}
        s_A(x1) = max{22, x1 + 5}
        times_A(x1,x2) = max{1, x1 - 3, x2 - 25}
        |0|_A = 0
  
    base order 2:
  
      precedence:
      
        plus# = |0| > s > times
      
  
    partial status:
  
      pi(plus#) = [2]
      pi(s) = []
      pi(times) = []
      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(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

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        p#_A(x1) = max{1, x1 - 5}
        s_A(x1) = max{7, x1 + 4}
        times_A(x1,x2) = max{5, x1 - 9, x2 - 14}
        |0|_A = 4
  
    base order 2:
  
      precedence:
      
        s > p# > times > |0|
      
  
    partial status:
  
      pi(p#) = []
      pi(s) = [1]
      pi(times) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.