Termination Proof using AProVE

Term Rewriting System R:
[x, y]
p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FACT(s(x)) -> *'(s(x), fact(p(s(x))))
FACT(s(x)) -> FACT(p(s(x)))
FACT(s(x)) -> P(s(x))
*'(s(x), y) -> +'(*(x, y), y)
*'(s(x), y) -> *'(x, y)
+'(x, s(y)) -> +'(x, y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
+'(x, s(y)) -> +'(x, y)

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
Dependency Pair:
*'(s(x), y) -> *'(x, y)

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
Dependency Pair:
FACT(s(x)) -> FACT(p(s(x)))

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
+'(x, s(y)) -> +'(x, y)

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
Dependency Pair:
*'(s(x), y) -> *'(x, y)

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
Dependency Pair:
FACT(s(x)) -> FACT(p(s(x)))

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
+'(x, s(y)) -> +'(x, y)

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
Dependency Pair:
*'(s(x), y) -> *'(x, y)

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
Dependency Pair:
FACT(s(x)) -> FACT(p(s(x)))

Rules:

p(s(x)) -> x
fact(0) -> s(0)
fact(s(x)) -> *(s(x), fact(p(s(x))))
*(0, y) -> 0
*(s(x), y) -> +(*(x, y), y)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Termination of R could not be shown.
Duration:
0:01 minutes