Term Rewriting System R:
[y, x]
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

LE(s(x), s(y)) -> LE(x, y)
MINUS(s(x), y) -> IFMINUS(le(s(x), y), s(x), y)
MINUS(s(x), y) -> LE(s(x), y)
IFMINUS(false, s(x), 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 four SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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

LE(s(x), s(y)) -> LE(x, y)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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

LE(s(s(x'')), s(s(y''))) -> LE(s(x''), s(y''))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳FwdInst`
`             ...`
`               →DP Problem 6`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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:

LE(s(s(s(x''''))), s(s(s(y'''')))) -> LE(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(LE(x1, x2)) =  1 + x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳FwdInst`
`             ...`
`               →DP Problem 7`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

IFMINUS(false, s(x), y) -> MINUS(x, y)
MINUS(s(x), y) -> IFMINUS(le(s(x), y), s(x), y)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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

MINUS(s(x), y) -> IFMINUS(le(s(x), y), s(x), y)
two new Dependency Pairs are created:

MINUS(s(x''), 0) -> IFMINUS(false, s(x''), 0)
MINUS(s(x''), s(y'')) -> IFMINUS(le(x'', y''), s(x''), s(y''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 8`
`             ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

MINUS(s(x''), s(y'')) -> IFMINUS(le(x'', y''), s(x''), s(y''))
MINUS(s(x''), 0) -> IFMINUS(false, s(x''), 0)
IFMINUS(false, s(x), y) -> MINUS(x, y)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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

MINUS(s(x''), s(y'')) -> IFMINUS(le(x'', y''), s(x''), s(y''))
three new Dependency Pairs are created:

MINUS(s(0), s(y''')) -> IFMINUS(true, s(0), s(y'''))
MINUS(s(s(x')), s(0)) -> IFMINUS(false, s(s(x')), s(0))
MINUS(s(s(x')), s(s(y'))) -> IFMINUS(le(x', y'), s(s(x')), s(s(y')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 8`
`             ↳Nar`
`             ...`
`               →DP Problem 9`
`                 ↳Instantiation Transformation`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

MINUS(s(s(x')), s(s(y'))) -> IFMINUS(le(x', y'), s(s(x')), s(s(y')))
MINUS(s(s(x')), s(0)) -> IFMINUS(false, s(s(x')), s(0))
IFMINUS(false, s(x), y) -> MINUS(x, y)
MINUS(s(x''), 0) -> IFMINUS(false, s(x''), 0)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

IFMINUS(false, s(x), y) -> MINUS(x, y)
three new Dependency Pairs are created:

IFMINUS(false, s(x'), 0) -> MINUS(x', 0)
IFMINUS(false, s(s(x''')), s(0)) -> MINUS(s(x'''), s(0))
IFMINUS(false, s(s(x'0')), s(s(y'''))) -> MINUS(s(x'0'), s(s(y''')))

The transformation is resulting in three new DP problems:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 8`
`             ↳Nar`
`             ...`
`               →DP Problem 10`
`                 ↳Forward Instantiation Transformation`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

IFMINUS(false, s(s(x'0')), s(s(y'''))) -> MINUS(s(x'0'), s(s(y''')))
MINUS(s(s(x')), s(s(y'))) -> IFMINUS(le(x', y'), s(s(x')), s(s(y')))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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

IFMINUS(false, s(s(x'0')), s(s(y'''))) -> MINUS(s(x'0'), s(s(y''')))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 8`
`             ↳Nar`
`             ...`
`               →DP Problem 13`
`                 ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

IFMINUS(false, s(s(s(x'''))), s(s(y''''))) -> MINUS(s(s(x''')), s(s(y'''')))
MINUS(s(s(x')), s(s(y'))) -> IFMINUS(le(x', y'), s(s(x')), s(s(y')))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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:

IFMINUS(false, s(s(s(x'''))), s(s(y''''))) -> MINUS(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(IF_MINUS(x1, x2, x3)) =  x2 POL(0) =  0 POL(false) =  0 POL(MINUS(x1, x2)) =  x1 POL(true) =  0 POL(s(x1)) =  1 + x1 POL(le(x1, x2)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 8`
`             ↳Nar`
`             ...`
`               →DP Problem 17`
`                 ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 8`
`             ↳Nar`
`             ...`
`               →DP Problem 11`
`                 ↳Forward Instantiation Transformation`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

IFMINUS(false, s(s(x''')), s(0)) -> MINUS(s(x'''), s(0))
MINUS(s(s(x')), s(0)) -> IFMINUS(false, s(s(x')), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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

IFMINUS(false, s(s(x''')), s(0)) -> MINUS(s(x'''), s(0))
one new Dependency Pair is created:

IFMINUS(false, s(s(s(x''''))), s(0)) -> MINUS(s(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 8`
`             ↳Nar`
`             ...`
`               →DP Problem 14`
`                 ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

IFMINUS(false, s(s(s(x''''))), s(0)) -> MINUS(s(s(x'''')), s(0))
MINUS(s(s(x')), s(0)) -> IFMINUS(false, s(s(x')), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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:

IFMINUS(false, s(s(s(x''''))), s(0)) -> MINUS(s(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(IF_MINUS(x1, x2, x3)) =  x2 POL(0) =  0 POL(false) =  0 POL(MINUS(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 8`
`             ↳Nar`
`             ...`
`               →DP Problem 18`
`                 ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

MINUS(s(s(x')), s(0)) -> IFMINUS(false, s(s(x')), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 8`
`             ↳Nar`
`             ...`
`               →DP Problem 12`
`                 ↳Forward Instantiation Transformation`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

IFMINUS(false, s(x'), 0) -> MINUS(x', 0)
MINUS(s(x''), 0) -> IFMINUS(false, s(x''), 0)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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

IFMINUS(false, s(x'), 0) -> MINUS(x', 0)
one new Dependency Pair is created:

IFMINUS(false, s(s(x'''')), 0) -> MINUS(s(x''''), 0)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 8`
`             ↳Nar`
`             ...`
`               →DP Problem 15`
`                 ↳Forward Instantiation Transformation`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

IFMINUS(false, s(s(x'''')), 0) -> MINUS(s(x''''), 0)
MINUS(s(x''), 0) -> IFMINUS(false, s(x''), 0)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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(s(x''), 0) -> IFMINUS(false, s(x''), 0)
one new Dependency Pair is created:

MINUS(s(s(x'''''')), 0) -> IFMINUS(false, s(s(x'''''')), 0)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 8`
`             ↳Nar`
`             ...`
`               →DP Problem 16`
`                 ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

MINUS(s(s(x'''''')), 0) -> IFMINUS(false, s(s(x'''''')), 0)
IFMINUS(false, s(s(x'''')), 0) -> MINUS(s(x''''), 0)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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:

IFMINUS(false, s(s(x'''')), 0) -> MINUS(s(x''''), 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(IF_MINUS(x1, x2, x3)) =  x2 POL(0) =  0 POL(false) =  0 POL(MINUS(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 8`
`             ↳Nar`
`             ...`
`               →DP Problem 19`
`                 ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

MINUS(s(s(x'''''')), 0) -> IFMINUS(false, s(s(x'''''')), 0)

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Narrowing Transformation`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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(0), s(y'')) -> QUOT(0, s(y''))
QUOT(s(s(x'')), s(y'')) -> QUOT(ifminus(le(s(x''), y''), s(x''), y''), s(y''))

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 20`
`             ↳Narrowing Transformation`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

QUOT(s(s(x'')), s(y'')) -> QUOT(ifminus(le(s(x''), y''), s(x''), y''), s(y''))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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(s(x'')), s(y'')) -> QUOT(ifminus(le(s(x''), y''), s(x''), y''), s(y''))
two new Dependency Pairs are created:

QUOT(s(s(x''')), s(0)) -> QUOT(ifminus(false, s(x'''), 0), s(0))
QUOT(s(s(x''')), s(s(y'))) -> QUOT(ifminus(le(x''', y'), s(x'''), s(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 3`
`         ↳Nar`
`           →DP Problem 20`
`             ↳Nar`
`             ...`
`               →DP Problem 21`
`                 ↳Rewriting Transformation`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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

QUOT(s(s(x''')), s(0)) -> QUOT(ifminus(false, s(x'''), 0), s(0))
one new Dependency Pair is created:

QUOT(s(s(x''')), s(0)) -> QUOT(s(minus(x''', 0)), 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 20`
`             ↳Nar`
`             ...`
`               →DP Problem 23`
`                 ↳Polynomial Ordering`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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(minus(x''', 0)), s(0))

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

minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(minus(x, y))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(QUOT(x1, x2)) =  1 + x1 POL(0) =  0 POL(false) =  0 POL(minus(x1, x2)) =  x1 POL(true) =  0 POL(s(x1)) =  1 + x1 POL(if_minus(x1, x2, x3)) =  x2 POL(le(x1, x2)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 20`
`             ↳Nar`
`             ...`
`               →DP Problem 22`
`                 ↳Polynomial Ordering`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

QUOT(s(s(x''')), s(s(y'))) -> QUOT(ifminus(le(x''', y'), s(x'''), s(y')), s(s(y')))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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(s(y'))) -> QUOT(ifminus(le(x''', y'), s(x'''), s(y')), s(s(y')))

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

minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(minus(x, y))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(QUOT(x1, x2)) =  x1 + x2 POL(0) =  0 POL(false) =  0 POL(minus(x1, x2)) =  x1 POL(true) =  0 POL(s(x1)) =  1 + x1 POL(if_minus(x1, x2, x3)) =  x2 POL(le(x1, x2)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 20`
`             ↳Nar`
`             ...`
`               →DP Problem 24`
`                 ↳Dependency Graph`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Narrowing Transformation`

Dependency Pair:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 26`
`             ↳Narrowing Transformation`

Dependency Pair:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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(s(x'')))) -> LOG(s(s(quot(minus(x'', s(0)), s(s(0))))))
two new Dependency Pairs are created:

LOG(s(s(s(0)))) -> LOG(s(s(quot(0, s(s(0))))))
LOG(s(s(s(s(x'))))) -> LOG(s(s(quot(ifminus(le(s(x'), s(0)), s(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 4`
`         ↳Nar`
`           →DP Problem 26`
`             ↳Nar`
`             ...`
`               →DP Problem 27`
`                 ↳Rewriting Transformation`

Dependency Pairs:

LOG(s(s(s(s(x'))))) -> LOG(s(s(quot(ifminus(le(s(x'), s(0)), s(x'), s(0)), s(s(0))))))
LOG(s(s(s(0)))) -> LOG(s(s(quot(0, s(s(0))))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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(0)))) -> LOG(s(s(quot(0, s(s(0))))))
one new Dependency Pair is created:

LOG(s(s(s(0)))) -> LOG(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 4`
`         ↳Nar`
`           →DP Problem 26`
`             ↳Nar`
`             ...`
`               →DP Problem 28`
`                 ↳Rewriting Transformation`

Dependency Pair:

LOG(s(s(s(s(x'))))) -> LOG(s(s(quot(ifminus(le(s(x'), s(0)), s(x'), s(0)), s(s(0))))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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(s(x'))))) -> LOG(s(s(quot(ifminus(le(s(x'), s(0)), s(x'), s(0)), s(s(0))))))
one new Dependency Pair is created:

LOG(s(s(s(s(x'))))) -> LOG(s(s(quot(ifminus(le(x', 0), s(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 4`
`         ↳Nar`
`           →DP Problem 26`
`             ↳Nar`
`             ...`
`               →DP Problem 29`
`                 ↳Polynomial Ordering`

Dependency Pair:

LOG(s(s(s(s(x'))))) -> LOG(s(s(quot(ifminus(le(x', 0), s(x'), s(0)), s(s(0))))))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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(s(x'))))) -> LOG(s(s(quot(ifminus(le(x', 0), s(x'), s(0)), s(s(0))))))

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

minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(minus(x, y))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(false) =  0 POL(minus(x1, x2)) =  x1 POL(quot(x1, x2)) =  x1 POL(true) =  0 POL(s(x1)) =  1 + x1 POL(if_minus(x1, x2, x3)) =  x2 POL(le(x1, x2)) =  0 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 4`
`         ↳Nar`
`           →DP Problem 26`
`             ↳Nar`
`             ...`
`               →DP Problem 30`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(0, y) -> 0
minus(s(x), y) -> ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) -> 0
ifminus(false, s(x), y) -> s(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:
0:01 minutes