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

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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

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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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 6

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳AFS

       →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

The following dependency pair can be strictly oriented:

-'(s(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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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 3

         ↳AFS

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 4

         ↳AFS

       →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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Argument Filtering and Ordering

       →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

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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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 3

         ↳AFS

       →DP Problem 4

         ↳AFS

           →DP Problem 9

             ↳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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 10

             ↳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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 11

                 ↳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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 12

                 ↳Argument Filtering and Ordering

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

The following dependency pairs can be strictly oriented:

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

The following usable rules for innermost w.r.t. to the AFS can be oriented:

ge(x, 0) -> true
ge(0, s(y)) -> false
ge(s(x), s(y)) -> ge(x, y)

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

{s, FACT} > {ge, true, false}
{s, FACT} > 0
{s, FACT} > IFFACT

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

FACT(x₁) -> FACT(x₁)
IFFACT(x₁, x₂) -> IFFACT(x₁, x₂)
s(x₁) -> s(x₁)
ge(x₁, x₂) -> ge(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 13

                 ↳Dependency Graph

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.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes