Termination Proof using AProVE

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.

     ↳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.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Remaining

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 AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

PLUS(x₁, x₂) -> PLUS(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Remaining

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.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Remaining

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 AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

P(x₁) -> P(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Remaining

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.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳Remaining

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 AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

TIMES(x₁, x₂) -> TIMES(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 4

         ↳Remaining

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.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:
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))

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