Term Rewriting System R:
[x, y]
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))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

PLUS(x, s(y)) -> PLUS(x, y)
TIMES(s(x), y) -> PLUS(times(x, y), y)
TIMES(s(x), y) -> TIMES(x, y)
P(s(s(x))) -> P(s(x))
FAC(s(x)) -> TIMES(fac(p(s(x))), s(x))
FAC(s(x)) -> FAC(p(s(x)))
FAC(s(x)) -> P(s(x))

Furthermore, R contains four SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Nar


Dependency Pair:

PLUS(x, s(y)) -> PLUS(x, y)


Rules:


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))





The following dependency pair can be strictly oriented:

PLUS(x, s(y)) -> PLUS(x, y)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(PLUS(x1, x2))=  x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Nar


Dependency Pair:


Rules:


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))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polynomial Ordering
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Nar


Dependency Pair:

P(s(s(x))) -> P(s(x))


Rules:


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))





The following dependency pair can be strictly oriented:

P(s(s(x))) -> P(s(x))


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(P(x1))=  1 + x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 6
             ↳Dependency Graph
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Nar


Dependency Pair:


Rules:


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))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polynomial Ordering
       →DP Problem 4
         ↳Nar


Dependency Pair:

TIMES(s(x), y) -> TIMES(x, y)


Rules:


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))





The following dependency pair can be strictly oriented:

TIMES(s(x), y) -> TIMES(x, y)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(TIMES(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 7
             ↳Dependency Graph
       →DP Problem 4
         ↳Nar


Dependency Pair:


Rules:


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))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Narrowing Transformation


Dependency Pair:

FAC(s(x)) -> FAC(p(s(x)))


Rules:


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))





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

FAC(s(x)) -> FAC(p(s(x)))
two new Dependency Pairs are created:

FAC(s(s(x''))) -> FAC(s(p(s(x''))))
FAC(s(0)) -> FAC(0)

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Nar
           →DP Problem 8
             ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

FAC(s(s(x''))) -> FAC(s(p(s(x''))))


Rules:


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))




Termination of R could not be shown.
Duration:
0:00 minutes