Termination Proof using AProVE

Term Rewriting System R:
[x, y]
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
ge(x, 0) -> true
ge(0, s(y)) -> false
ge(s(x), s(y)) -> ge(x, y)
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
fact(x) -> iffact(x, ge(x, s(s(0))))
iffact(x, true) -> *(x, fact(-(x, s(0))))
iffact(x, false) -> s(0)

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

+'(x, s(y)) -> +'(x, y)
*'(x, s(y)) -> +'(*(x, y), x)
*'(x, s(y)) -> *'(x, y)
GE(s(x), s(y)) -> GE(x, y)
-'(s(x), s(y)) -> -'(x, y)
FACT(x) -> IFFACT(x, ge(x, s(s(0))))
FACT(x) -> GE(x, s(s(0)))
IFFACT(x, true) -> *'(x, fact(-(x, s(0))))
IFFACT(x, true) -> FACT(-(x, s(0)))
IFFACT(x, true) -> -'(x, s(0))

Furthermore, R contains five SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

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

Rules:

+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
ge(x, 0) -> true
ge(0, s(y)) -> false
ge(s(x), s(y)) -> ge(x, y)
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
fact(x) -> iffact(x, ge(x, s(s(0))))
iffact(x, true) -> *(x, fact(-(x, s(0))))
iffact(x, false) -> s(0)

The following dependency pair can be strictly oriented:

+'(x, s(y)) -> +'(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(s(x₁)) = 1 + x₁

POL(+'(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
ge(x, 0) -> true
ge(0, s(y)) -> false
ge(s(x), s(y)) -> ge(x, y)
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
fact(x) -> iffact(x, ge(x, s(s(0))))
iffact(x, true) -> *(x, fact(-(x, s(0))))
iffact(x, false) -> s(0)

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

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

Rules:

+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
ge(x, 0) -> true
ge(0, s(y)) -> false
ge(s(x), s(y)) -> ge(x, y)
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
fact(x) -> iffact(x, ge(x, s(s(0))))
iffact(x, true) -> *(x, fact(-(x, s(0))))
iffact(x, false) -> s(0)

The following dependency pair can be strictly oriented:

GE(s(x), s(y)) -> GE(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(GE(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
ge(x, 0) -> true
ge(0, s(y)) -> false
ge(s(x), s(y)) -> ge(x, y)
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
fact(x) -> iffact(x, ge(x, s(s(0))))
iffact(x, true) -> *(x, fact(-(x, s(0))))
iffact(x, false) -> s(0)

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polynomial Ordering

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

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

Rules:

+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
ge(x, 0) -> true
ge(0, s(y)) -> false
ge(s(x), s(y)) -> ge(x, y)
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
fact(x) -> iffact(x, ge(x, s(s(0))))
iffact(x, true) -> *(x, fact(-(x, s(0))))
iffact(x, false) -> s(0)

The following dependency pair can be strictly oriented:

-'(s(x), s(y)) -> -'(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(-'(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
ge(x, 0) -> true
ge(0, s(y)) -> false
ge(s(x), s(y)) -> ge(x, y)
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
fact(x) -> iffact(x, ge(x, s(s(0))))
iffact(x, true) -> *(x, fact(-(x, s(0))))
iffact(x, false) -> s(0)

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polynomial Ordering

       →DP Problem 5

         ↳Remaining

Dependency Pair:

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

Rules:

+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
ge(x, 0) -> true
ge(0, s(y)) -> false
ge(s(x), s(y)) -> ge(x, y)
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
fact(x) -> iffact(x, ge(x, s(s(0))))
iffact(x, true) -> *(x, fact(-(x, s(0))))
iffact(x, false) -> s(0)

The following dependency pair can be strictly oriented:

*'(x, s(y)) -> *'(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(*'(x₁, x₂)) = x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

           →DP Problem 9

             ↳Dependency Graph

       →DP Problem 5

         ↳Remaining

Dependency Pair:

Rules:

+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
ge(x, 0) -> true
ge(0, s(y)) -> false
ge(s(x), s(y)) -> ge(x, y)
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
fact(x) -> iffact(x, ge(x, s(s(0))))
iffact(x, true) -> *(x, fact(-(x, s(0))))
iffact(x, false) -> s(0)

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

IFFACT(x, true) -> FACT(-(x, s(0)))
FACT(x) -> IFFACT(x, ge(x, s(s(0))))

Rules:

+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
ge(x, 0) -> true
ge(0, s(y)) -> false
ge(s(x), s(y)) -> ge(x, y)
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
fact(x) -> iffact(x, ge(x, s(s(0))))
iffact(x, true) -> *(x, fact(-(x, s(0))))
iffact(x, false) -> s(0)

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