Termination Proof using AProVE

Term Rewriting System R:
[x, y]
fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
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)

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FAC(0) -> 1'
FAC(s(x)) -> *'(s(x), fac(x))
FAC(s(x)) -> FAC(x)
FLOOP(s(x), y) -> FLOOP(x, *(s(x), y))
FLOOP(s(x), y) -> *'(s(x), y)
*'(x, s(y)) -> +'(*(x, y), x)
*'(x, s(y)) -> *'(x, y)
+'(x, s(y)) -> +'(x, y)

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

Dependency Pair:

+'(x, s(y)) -> +'(x, y)

Rules:

fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
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)

Strategy:

innermost

The following dependency pair can be strictly oriented:

+'(x, s(y)) -> +'(x, y)

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(+'(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.
Used Argument Filtering System:

+'(x₁, x₂) -> +'(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

         ↳AFS

Dependency Pair:

Rules:

fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
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)

Strategy:

innermost

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

         ↳AFS

Dependency Pair:

*'(x, s(y)) -> *'(x, y)

Rules:

fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
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)

Strategy:

innermost

The following dependency pair can be strictly oriented:

*'(x, s(y)) -> *'(x, y)

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*'(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

*'(x₁, x₂) -> *'(x₁, x₂)
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

         ↳AFS

Dependency Pair:

Rules:

fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
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)

Strategy:

innermost

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

         ↳AFS

Dependency Pair:

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

Rules:

fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
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)

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(FAC(x₁)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

FAC(x₁) -> FAC(x₁)
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

         ↳AFS

Dependency Pair:

Rules:

fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
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)

Strategy:

innermost

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

         ↳Argument Filtering and Ordering

Dependency Pair:

FLOOP(s(x), y) -> FLOOP(x, *(s(x), y))

Rules:

fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
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)

Strategy:

innermost

The following dependency pair can be strictly oriented:

FLOOP(s(x), y) -> FLOOP(x, *(s(x), y))

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

FLOOP(x₁, x₂) -> x₁
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

           →DP Problem 8

             ↳Dependency Graph

Dependency Pair:

Rules:

fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
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)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes

POL(s(x₁))	= 1 + x₁
POL(+'(x₁, x₂))	= x₁ + x₂