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)

Innermost 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

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

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)

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 6

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 7

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(+'(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 8

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

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)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

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)

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 9

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 9

             ↳FwdInst

...

               →DP Problem 10

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(GE(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 9

             ↳FwdInst

...

               →DP Problem 11

                 ↳Dependency Graph

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

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)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

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)

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 12

             ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 12

             ↳FwdInst

...

               →DP Problem 13

                 ↳Polynomial Ordering

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(-'(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 12

             ↳FwdInst

...

               →DP Problem 14

                 ↳Dependency Graph

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

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)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳Nar

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)

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 15

             ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pair:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 15

             ↳FwdInst

...

               →DP Problem 16

                 ↳Polynomial Ordering

       →DP Problem 5

         ↳Nar

Dependency Pair:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*'(x₁, x₂)) = 1 + x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 15

             ↳FwdInst

...

               →DP Problem 17

                 ↳Dependency Graph

       →DP Problem 5

         ↳Nar

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)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Narrowing Transformation

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)

Strategy:

innermost

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

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

two new Dependency Pairs are created:

FACT(0) -> IFFACT(0, false)
FACT(s(x'')) -> IFFACT(s(x''), ge(x'', s(0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Narrowing Transformation

Dependency Pairs:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 19

                 ↳Rewriting Transformation

Dependency Pairs:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

IFFACT(s(x''), true) -> FACT(x'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 20

                 ↳Narrowing Transformation

Dependency Pairs:

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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

FACT(s(0)) -> IFFACT(s(0), false)
FACT(s(s(x'))) -> IFFACT(s(s(x')), ge(x', 0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 21

                 ↳Rewriting Transformation

Dependency Pairs:

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

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)

Strategy:

innermost

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

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

one new Dependency Pair is created:

FACT(s(s(x'))) -> IFFACT(s(s(x')), true)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 22

                 ↳Polynomial Ordering

Dependency Pairs:

FACT(s(s(x'))) -> IFFACT(s(s(x')), true)
IFFACT(s(x''), true) -> FACT(x'')

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)

Strategy:

innermost

The following dependency pair can be strictly oriented:

IFFACT(s(x''), true) -> FACT(x'')

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(FACT(x₁)) = x₁

POL(IFFACT(x₁, x₂)) = x₁

POL(true) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 23

                 ↳Dependency Graph

Dependency Pair:

FACT(s(s(x'))) -> IFFACT(s(s(x')), true)

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)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(FACT(x₁))	= x₁
POL(IFFACT(x₁, x₂))	= x₁
POL(true)	= 0
POL(s(x₁))	= 1 + x₁