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

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Remaining

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

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 5

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Remaining

Dependency Pair:

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

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 5

             ↳FwdInst

...

               →DP Problem 6

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Remaining

Dependency Pair:

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

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

+'(x₁, x₂) -> +'(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 7

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Remaining

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

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Remaining

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

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 8

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Remaining

Dependency Pair:

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

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 8

             ↳FwdInst

...

               →DP Problem 9

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Remaining

Dependency Pair:

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

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 8

             ↳FwdInst

...

               →DP Problem 10

                 ↳Dependency Graph

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Remaining

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

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Remaining

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

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

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

one new Dependency Pair is created:

FAC(s(s(x''))) -> FAC(s(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 11

             ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

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

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

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

one new Dependency Pair is created:

FAC(s(s(s(x'''')))) -> FAC(s(s(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 11

             ↳FwdInst

...

               →DP Problem 12

                 ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳Remaining

Dependency Pair:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(FAC(x₁)) = 1 + 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

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 11

             ↳FwdInst

...

               →DP Problem 13

                 ↳Dependency Graph

       →DP Problem 4

         ↳Remaining

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

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:
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

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

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

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