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.

R
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.

R
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:

R
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:

R
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 w.r.t. to the implicit AFS 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.

R
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.

R
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:

R
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:

R
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 w.r.t. to the implicit AFS 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.

R
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.

R
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:

R
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 w.r.t. to the implicit AFS 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.

R
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 w.r.t. to the implicit AFS 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 + x2 POL(0) =  1 POL(min(x1, x2)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

R
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:

R
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:

R
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:

R
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 w.r.t. to the implicit AFS 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.

R
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