Termination Proof using AProVE

Term Rewriting System R:
[X, Y]
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MIN(s(X), s(Y)) -> MIN(X, Y)
QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MIN(X, Y)
LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))
LOG(s(s(X))) -> QUOT(X, s(s(0)))

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pair:

MIN(s(X), s(Y)) -> MIN(X, Y)

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

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

MIN(s(X), s(Y)) -> MIN(X, Y)

one new Dependency Pair is created:

MIN(s(s(X'')), s(s(Y''))) -> MIN(s(X''), s(Y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pair:

MIN(s(s(X'')), s(s(Y''))) -> MIN(s(X''), s(Y''))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

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

MIN(s(s(X'')), s(s(Y''))) -> MIN(s(X''), s(Y''))

one new Dependency Pair is created:

MIN(s(s(s(X''''))), s(s(s(Y'''')))) -> MIN(s(s(X'''')), s(s(Y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pair:

MIN(s(s(s(X''''))), s(s(s(Y'''')))) -> MIN(s(s(X'''')), s(s(Y'''')))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MIN(s(s(s(X''''))), s(s(s(Y'''')))) -> MIN(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(MIN(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Narrowing Transformation

       →DP Problem 3

         ↳Nar

Dependency Pair:

QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

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

QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))

two new Dependency Pairs are created:

QUOT(s(X''), s(0)) -> QUOT(X'', s(0))
QUOT(s(s(X'')), s(s(Y''))) -> QUOT(min(X'', Y''), s(s(Y'')))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳Forward Instantiation Transformation

           →DP Problem 8

             ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pair:

QUOT(s(X''), s(0)) -> QUOT(X'', s(0))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

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

QUOT(s(X''), s(0)) -> QUOT(X'', s(0))

one new Dependency Pair is created:

QUOT(s(s(X'''')), s(0)) -> QUOT(s(X''''), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Polynomial Ordering

           →DP Problem 8

             ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pair:

QUOT(s(s(X'''')), s(0)) -> QUOT(s(X''''), s(0))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

QUOT(s(s(X'''')), s(0)) -> QUOT(s(X''''), s(0))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = 1 + x₁

POL(0) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 12

                 ↳Dependency Graph

           →DP Problem 8

             ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳FwdInst

           →DP Problem 8

             ↳Narrowing Transformation

       →DP Problem 3

         ↳Nar

Dependency Pair:

QUOT(s(s(X'')), s(s(Y''))) -> QUOT(min(X'', Y''), s(s(Y'')))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

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

QUOT(s(s(X'')), s(s(Y''))) -> QUOT(min(X'', Y''), s(s(Y'')))

two new Dependency Pairs are created:

QUOT(s(s(X''')), s(s(0))) -> QUOT(X''', s(s(0)))
QUOT(s(s(s(X'))), s(s(s(Y')))) -> QUOT(min(X', Y'), s(s(s(Y'))))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳FwdInst

           →DP Problem 8

             ↳Nar

...

               →DP Problem 10

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

QUOT(s(s(X''')), s(s(0))) -> QUOT(X''', s(s(0)))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

QUOT(s(s(X''')), s(s(0))) -> QUOT(X''', s(s(0)))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = 1 + x₁

POL(0) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳FwdInst

           →DP Problem 8

             ↳Nar

...

               →DP Problem 11

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

QUOT(s(s(s(X'))), s(s(s(Y')))) -> QUOT(min(X', Y'), s(s(s(Y'))))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

QUOT(s(s(s(X'))), s(s(s(Y')))) -> QUOT(min(X', Y'), s(s(s(Y'))))

Additionally, the following usable rules for innermost can be oriented:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = 1 + x₁

POL(0) = 1

POL(min(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Narrowing Transformation

Dependency Pair:

LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

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

LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))

two new Dependency Pairs are created:

LOG(s(s(0))) -> LOG(s(0))
LOG(s(s(s(X'')))) -> LOG(s(s(quot(min(X'', s(0)), s(s(0))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 15

             ↳Narrowing Transformation

Dependency Pair:

LOG(s(s(s(X'')))) -> LOG(s(s(quot(min(X'', s(0)), s(s(0))))))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

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

LOG(s(s(s(X'')))) -> LOG(s(s(quot(min(X'', s(0)), s(s(0))))))

one new Dependency Pair is created:

LOG(s(s(s(s(X'))))) -> LOG(s(s(quot(min(X', 0), s(s(0))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 15

             ↳Nar

...

               →DP Problem 16

                 ↳Rewriting Transformation

Dependency Pair:

LOG(s(s(s(s(X'))))) -> LOG(s(s(quot(min(X', 0), s(s(0))))))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

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

LOG(s(s(s(s(X'))))) -> LOG(s(s(quot(min(X', 0), s(s(0))))))

one new Dependency Pair is created:

LOG(s(s(s(s(X'))))) -> LOG(s(s(quot(X', s(s(0))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 15

             ↳Nar

...

               →DP Problem 17

                 ↳Polynomial Ordering

Dependency Pair:

LOG(s(s(s(s(X'))))) -> LOG(s(s(quot(X', s(s(0))))))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

LOG(s(s(s(s(X'))))) -> LOG(s(s(quot(X', s(s(0))))))

Additionally, the following usable rules for innermost can be oriented:

quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(quot(x₁, x₂)) = x₁

POL(min(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

POL(LOG(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 15

             ↳Nar

...

               →DP Problem 18

                 ↳Dependency Graph

Dependency Pair:

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(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(MIN(x₁, x₂))	= 1 + x₁
POL(s(x₁))	= 1 + x₁

POL(QUOT(x₁, x₂))	= 1 + x₁
POL(0)	= 0
POL(s(x₁))	= 1 + x₁

POL(0)	= 0
POL(quot(x₁, x₂))	= x₁
POL(min(x₁, x₂))	= x₁
POL(s(x₁))	= 1 + x₁
POL(LOG(x₁))	= 1 + x₁