Termination Proof using AProVE

Term Rewriting System R:
[x, y]
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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:

MINUS(x, s(y)) -> PRED(minus(x, y))
MINUS(x, s(y)) -> MINUS(x, y)
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) -> MINUS(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:

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

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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

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

one new Dependency Pair is created:

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

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

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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

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

one new Dependency Pair is created:

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

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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:

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

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MINUS(x₁, x₂)) = 1 + x₁ + x₂

POL(s(x₁)) = 1 + x₁

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

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

     ↳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:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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(minus(x, y), s(y))

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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(minus(x, y), s(y))

two new Dependency Pairs are created:

QUOT(s(x''), s(0)) -> QUOT(x'', s(0))
QUOT(s(x''), s(s(y''))) -> QUOT(pred(minus(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:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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

                 ↳Argument Filtering and Ordering

           →DP Problem 8

             ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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 w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = 1 + x₁ + x₂

POL(0) = 0

POL(s(x₁)) = 1 + x₁

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

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

     ↳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:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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(x''), s(s(y''))) -> QUOT(pred(minus(x'', y'')), s(s(y'')))

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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(s(y''))) -> QUOT(pred(minus(x'', y'')), s(s(y'')))

two new Dependency Pairs are created:

QUOT(s(x'''), s(s(0))) -> QUOT(pred(x'''), s(s(0)))
QUOT(s(x'''), s(s(s(y')))) -> QUOT(pred(pred(minus(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

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

QUOT(s(x'''), s(s(0))) -> QUOT(pred(x'''), s(s(0)))

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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(x'''), s(s(0))) -> QUOT(pred(x'''), s(s(0)))

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

pred(s(x)) -> x

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = x₁ + x₂

POL(0) = 0

POL(pred(x₁)) = x₁

POL(s(x₁)) = 1 + x₁

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

QUOT(x₁, x₂) -> QUOT(x₁, x₂)
s(x₁) -> s(x₁)
pred(x₁) -> pred(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 7

             ↳FwdInst

           →DP Problem 8

             ↳Nar

...

               →DP Problem 11

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

QUOT(s(x'''), s(s(s(y')))) -> QUOT(pred(pred(minus(x''', y'))), s(s(s(y'))))

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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(x'''), s(s(s(y')))) -> QUOT(pred(pred(minus(x''', y'))), s(s(s(y'))))

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

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = x₁ + x₂

POL(pred(x₁)) = x₁

POL(s(x₁)) = 1 + x₁

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

QUOT(x₁, x₂) -> QUOT(x₁, x₂)
s(x₁) -> s(x₁)
pred(x₁) -> pred(x₁)
minus(x₁, x₂) -> x₁

     ↳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:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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(minus(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

             ↳Rewriting Transformation

Dependency Pair:

LOG(s(s(s(x'')))) -> LOG(s(s(quot(minus(x'', s(0)), s(s(0))))))

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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(x'')))) -> LOG(s(s(quot(minus(x'', s(0)), s(s(0))))))

one new Dependency Pair is created:

LOG(s(s(s(x'')))) -> LOG(s(s(quot(pred(minus(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

             ↳Rw

...

               →DP Problem 16

                 ↳Rewriting Transformation

Dependency Pair:

LOG(s(s(s(x'')))) -> LOG(s(s(quot(pred(minus(x'', 0)), s(s(0))))))

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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(x'')))) -> LOG(s(s(quot(pred(minus(x'', 0)), s(s(0))))))

one new Dependency Pair is created:

LOG(s(s(s(x'')))) -> LOG(s(s(quot(pred(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

             ↳Rw

...

               →DP Problem 17

                 ↳Argument Filtering and Ordering

Dependency Pair:

LOG(s(s(s(x'')))) -> LOG(s(s(quot(pred(x''), s(s(0))))))

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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(x'')))) -> LOG(s(s(quot(pred(x''), s(s(0))))))

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

pred(s(x)) -> x
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(pred(x₁)) = x₁

POL(s(x₁)) = 1 + x₁

POL(LOG(x₁)) = 1 + x₁

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

LOG(x₁) -> LOG(x₁)
s(x₁) -> s(x₁)
quot(x₁, x₂) -> x₁
pred(x₁) -> pred(x₁)
minus(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 15

             ↳Rw

...

               →DP Problem 18

                 ↳Dependency Graph

Dependency Pair:

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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(MINUS(x₁, x₂))	= 1 + x₁ + x₂
POL(s(x₁))	= 1 + x₁

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

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