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

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

TIMES(x, plus(y, s(z))) -> PLUS(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
TIMES(x, plus(y, s(z))) -> TIMES(x, plus(y, times(s(z), 0)))
TIMES(x, plus(y, s(z))) -> PLUS(y, times(s(z), 0))
TIMES(x, plus(y, s(z))) -> TIMES(s(z), 0)
TIMES(x, plus(y, s(z))) -> TIMES(x, s(z))
TIMES(x, s(y)) -> PLUS(times(x, y), x)
TIMES(x, s(y)) -> TIMES(x, y)
PLUS(x, s(y)) -> PLUS(x, y)

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳FwdInst`

Dependency Pair:

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

Rules:

times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))

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:

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

Dependency Pair:

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

Rules:

times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))

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:

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

Dependency Pair:

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

Rules:

times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))

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 w.r.t. to the implicit AFS 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.

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

Dependency Pair:

Rules:

times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Forward Instantiation Transformation`

Dependency Pair:

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

Rules:

times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))

Strategy:

innermost

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

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳Forward Instantiation Transformation`

Dependency Pair:

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

Rules:

times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))

Strategy:

innermost

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

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

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 7`
`                 ↳Polynomial Ordering`

Dependency Pair:

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

Rules:

times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 6`
`             ↳FwdInst`
`             ...`
`               →DP Problem 8`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes