Termination Proof using AProVE

Term Rewriting System R:
[x, y]
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

PLUS(x, s(y)) -> PLUS(x, y)
TIMES(s(x), y) -> PLUS(times(x, y), y)
TIMES(s(x), y) -> TIMES(x, y)
P(s(s(x))) -> P(s(x))
FAC(s(x)) -> TIMES(fac(p(s(x))), s(x))
FAC(s(x)) -> FAC(p(s(x)))
FAC(s(x)) -> P(s(x))

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

         ↳Nar

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(PLUS(x₁, x₂)) = 1 + x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳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

         ↳Nar

Dependency Pair:

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

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

         ↳Nar

Dependency Pair:

P(s(s(x))) -> P(s(x))

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Strategy:

innermost

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

P(s(s(x))) -> P(s(x))

one new Dependency Pair is created:

P(s(s(s(x'')))) -> P(s(s(x'')))

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

         ↳Nar

Dependency Pair:

P(s(s(s(x'')))) -> P(s(s(x'')))

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Strategy:

innermost

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

P(s(s(s(x'')))) -> P(s(s(x'')))

one new Dependency Pair is created:

P(s(s(s(s(x''''))))) -> P(s(s(s(x''''))))

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

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Nar

Dependency Pair:

P(s(s(s(s(x''''))))) -> P(s(s(s(x''''))))

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

P(s(s(s(s(x''''))))) -> P(s(s(s(x''''))))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(P(x₁)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳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

         ↳Nar

Dependency Pair:

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

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

         ↳Nar

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Strategy:

innermost

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

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

one new Dependency Pair is created:

TIMES(s(s(x'')), y'') -> TIMES(s(x''), 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 11

             ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Strategy:

innermost

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

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

one new Dependency Pair is created:

TIMES(s(s(s(x''''))), y'''') -> TIMES(s(s(x'''')), 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 11

             ↳FwdInst

...

               →DP Problem 12

                 ↳Polynomial Ordering

       →DP Problem 4

         ↳Nar

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(TIMES(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 11

             ↳FwdInst

...

               →DP Problem 13

                 ↳Dependency Graph

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

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

         ↳Narrowing Transformation

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Strategy:

innermost

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

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

two new Dependency Pairs are created:

FAC(s(s(x''))) -> FAC(s(p(s(x''))))
FAC(s(0)) -> FAC(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

         ↳Nar

           →DP Problem 14

             ↳Narrowing Transformation

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Strategy:

innermost

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

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

two new Dependency Pairs are created:

FAC(s(s(s(x')))) -> FAC(s(s(p(s(x')))))
FAC(s(s(0))) -> FAC(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

         ↳Nar

           →DP Problem 14

             ↳Nar

...

               →DP Problem 15

                 ↳Narrowing Transformation

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Strategy:

innermost

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

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

two new Dependency Pairs are created:

FAC(s(s(s(s(x''))))) -> FAC(s(s(s(p(s(x''))))))
FAC(s(s(s(0)))) -> FAC(s(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

         ↳Nar

           →DP Problem 14

             ↳Nar

...

               →DP Problem 16

                 ↳Narrowing Transformation

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Strategy:

innermost

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

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

two new Dependency Pairs are created:

FAC(s(s(s(s(s(x')))))) -> FAC(s(s(s(s(p(s(x')))))))
FAC(s(s(s(s(0))))) -> FAC(s(s(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

         ↳Nar

           →DP Problem 14

             ↳Nar

...

               →DP Problem 17

                 ↳Narrowing Transformation

Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Strategy:

innermost

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

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

two new Dependency Pairs are created:

FAC(s(s(s(s(s(s(x''))))))) -> FAC(s(s(s(s(s(p(s(x''))))))))
FAC(s(s(s(s(s(0)))))) -> FAC(s(s(s(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

         ↳Nar

           →DP Problem 14

             ↳Nar

...

               →DP Problem 18

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

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

Rules:

plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Strategy:

innermost

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

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

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