YES

We show the termination of the TRS R:

  fac(|0|()) -> |1|()
  fac(s(x)) -> *(s(x),fac(x))
  floop(|0|(),y) -> y
  floop(s(x),y) -> floop(x,*(s(x),y))
  *(x,|0|()) -> |0|()
  *(x,s(y)) -> +(*(x,y),x)
  +(x,|0|()) -> x
  +(x,s(y)) -> s(+(x,y))
  |1|() -> s(|0|())
  fac(|0|()) -> s(|0|())

-- SCC decomposition.

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

p1: fac#(|0|()) -> |1|#()
p2: fac#(s(x)) -> *#(s(x),fac(x))
p3: fac#(s(x)) -> fac#(x)
p4: floop#(s(x),y) -> floop#(x,*(s(x),y))
p5: floop#(s(x),y) -> *#(s(x),y)
p6: *#(x,s(y)) -> +#(*(x,y),x)
p7: *#(x,s(y)) -> *#(x,y)
p8: +#(x,s(y)) -> +#(x,y)

and R consists of:

r1: fac(|0|()) -> |1|()
r2: fac(s(x)) -> *(s(x),fac(x))
r3: floop(|0|(),y) -> y
r4: floop(s(x),y) -> floop(x,*(s(x),y))
r5: *(x,|0|()) -> |0|()
r6: *(x,s(y)) -> +(*(x,y),x)
r7: +(x,|0|()) -> x
r8: +(x,s(y)) -> s(+(x,y))
r9: |1|() -> s(|0|())
r10: fac(|0|()) -> s(|0|())

The estimated dependency graph contains the following SCCs:

  {p3}
  {p4}
  {p7}
  {p8}


-- Reduction pair.

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

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

and R consists of:

r1: fac(|0|()) -> |1|()
r2: fac(s(x)) -> *(s(x),fac(x))
r3: floop(|0|(),y) -> y
r4: floop(s(x),y) -> floop(x,*(s(x),y))
r5: *(x,|0|()) -> |0|()
r6: *(x,s(y)) -> +(*(x,y),x)
r7: +(x,|0|()) -> x
r8: +(x,s(y)) -> s(+(x,y))
r9: |1|() -> s(|0|())
r10: fac(|0|()) -> s(|0|())

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        fac# > s
      
      argument filter:
    
        pi(fac#) = 1
        pi(s) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        fac#_A(x1) = x1
        s_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

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: floop#(s(x),y) -> floop#(x,*(s(x),y))

and R consists of:

r1: fac(|0|()) -> |1|()
r2: fac(s(x)) -> *(s(x),fac(x))
r3: floop(|0|(),y) -> y
r4: floop(s(x),y) -> floop(x,*(s(x),y))
r5: *(x,|0|()) -> |0|()
r6: *(x,s(y)) -> +(*(x,y),x)
r7: +(x,|0|()) -> x
r8: +(x,s(y)) -> s(+(x,y))
r9: |1|() -> s(|0|())
r10: fac(|0|()) -> s(|0|())

The set of usable rules consists of

  r5, r6, r7, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        |0| > floop# > * > + > s
      
      argument filter:
    
        pi(floop#) = [1, 2]
        pi(s) = [1]
        pi(*) = [1, 2]
        pi(+) = [1, 2]
        pi(|0|) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        floop#_A(x1,x2) = 0
        s_A(x1) = 2
        *_A(x1,x2) = x1 + 1
        +_A(x1,x2) = x1 + 1
        |0|_A() = 2
    

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: *#(x,s(y)) -> *#(x,y)

and R consists of:

r1: fac(|0|()) -> |1|()
r2: fac(s(x)) -> *(s(x),fac(x))
r3: floop(|0|(),y) -> y
r4: floop(s(x),y) -> floop(x,*(s(x),y))
r5: *(x,|0|()) -> |0|()
r6: *(x,s(y)) -> +(*(x,y),x)
r7: +(x,|0|()) -> x
r8: +(x,s(y)) -> s(+(x,y))
r9: |1|() -> s(|0|())
r10: fac(|0|()) -> s(|0|())

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > *#
      
      argument filter:
    
        pi(*#) = 2
        pi(s) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        *#_A(x1,x2) = x2
        s_A(x1) = x1 + 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: +#(x,s(y)) -> +#(x,y)

and R consists of:

r1: fac(|0|()) -> |1|()
r2: fac(s(x)) -> *(s(x),fac(x))
r3: floop(|0|(),y) -> y
r4: floop(s(x),y) -> floop(x,*(s(x),y))
r5: *(x,|0|()) -> |0|()
r6: *(x,s(y)) -> +(*(x,y),x)
r7: +(x,|0|()) -> x
r8: +(x,s(y)) -> s(+(x,y))
r9: |1|() -> s(|0|())
r10: fac(|0|()) -> s(|0|())

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > +#
      
      argument filter:
    
        pi(+#) = 2
        pi(s) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        +#_A(x1,x2) = x2
        s_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1

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