Term Rewriting System R:
[n, x, m, y]
sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

SUM(cons(s(n), x), cons(m, y)) -> SUM(cons(n, x), cons(s(m), y))
SUM(cons(0, x), y) -> SUM(x, y)
WEIGHT(cons(n, cons(m, x))) -> WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))
WEIGHT(cons(n, cons(m, x))) -> SUM(cons(n, cons(m, x)), cons(0, x))

Furthermore, R contains two SCCs.

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

Dependency Pairs:

SUM(cons(0, x), y) -> SUM(x, y)
SUM(cons(s(n), x), cons(m, y)) -> SUM(cons(n, x), cons(s(m), y))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

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

SUM(cons(s(n), x), cons(m, y)) -> SUM(cons(n, x), cons(s(m), y))
two new Dependency Pairs are created:

SUM(cons(s(s(n'')), x''), cons(m'', y'')) -> SUM(cons(s(n''), x''), cons(s(m''), y''))
SUM(cons(s(0), x''), cons(m', y'')) -> SUM(cons(0, x''), cons(s(m'), 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`
`         ↳Nar`

Dependency Pairs:

SUM(cons(s(0), x''), cons(m', y'')) -> SUM(cons(0, x''), cons(s(m'), y''))
SUM(cons(s(s(n'')), x''), cons(m'', y'')) -> SUM(cons(s(n''), x''), cons(s(m''), y''))
SUM(cons(0, x), y) -> SUM(x, y)

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

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

SUM(cons(0, x), y) -> SUM(x, y)
three new Dependency Pairs are created:

SUM(cons(0, cons(0, x'')), y'') -> SUM(cons(0, x''), y'')
SUM(cons(0, cons(s(s(n'''')), x'''')), cons(m'''', y'''')) -> SUM(cons(s(s(n'''')), x''''), cons(m'''', y''''))
SUM(cons(0, cons(s(0), x'''')), cons(m''', y'''')) -> SUM(cons(s(0), x''''), cons(m''', y''''))

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

SUM(cons(0, cons(s(0), x'''')), cons(m''', y'''')) -> SUM(cons(s(0), x''''), cons(m''', y''''))
SUM(cons(s(s(n'')), x''), cons(m'', y'')) -> SUM(cons(s(n''), x''), cons(s(m''), y''))
SUM(cons(0, cons(s(s(n'''')), x'''')), cons(m'''', y'''')) -> SUM(cons(s(s(n'''')), x''''), cons(m'''', y''''))
SUM(cons(0, cons(0, x'')), y'') -> SUM(cons(0, x''), y'')
SUM(cons(s(0), x''), cons(m', y'')) -> SUM(cons(0, x''), cons(s(m'), y''))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

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

SUM(cons(s(s(n'')), x''), cons(m'', y'')) -> SUM(cons(s(n''), x''), cons(s(m''), y''))
two new Dependency Pairs are created:

SUM(cons(s(s(s(n''''))), x''''), cons(m'''', y'''')) -> SUM(cons(s(s(n'''')), x''''), cons(s(m''''), y''''))
SUM(cons(s(s(0)), x''''), cons(m'''', y'''')) -> SUM(cons(s(0), x''''), cons(s(m''''), y''''))

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

SUM(cons(s(s(0)), x''''), cons(m'''', y'''')) -> SUM(cons(s(0), x''''), cons(s(m''''), y''''))
SUM(cons(s(s(s(n''''))), x''''), cons(m'''', y'''')) -> SUM(cons(s(s(n'''')), x''''), cons(s(m''''), y''''))
SUM(cons(0, cons(s(s(n'''')), x'''')), cons(m'''', y'''')) -> SUM(cons(s(s(n'''')), x''''), cons(m'''', y''''))
SUM(cons(0, cons(0, x'')), y'') -> SUM(cons(0, x''), y'')
SUM(cons(s(0), x''), cons(m', y'')) -> SUM(cons(0, x''), cons(s(m'), y''))
SUM(cons(0, cons(s(0), x'''')), cons(m''', y'''')) -> SUM(cons(s(0), x''''), cons(m''', y''''))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

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

SUM(cons(s(0), x''), cons(m', y'')) -> SUM(cons(0, x''), cons(s(m'), y''))
three new Dependency Pairs are created:

SUM(cons(s(0), cons(0, x'''')), cons(m'', y'''')) -> SUM(cons(0, cons(0, x'''')), cons(s(m''), y''''))
SUM(cons(s(0), cons(s(s(n'''''')), x'''''')), cons(m'', y''')) -> SUM(cons(0, cons(s(s(n'''''')), x'''''')), cons(s(m''), y'''))
SUM(cons(s(0), cons(s(0), x'''''')), cons(m'', y''')) -> SUM(cons(0, cons(s(0), x'''''')), cons(s(m''), y'''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 6`
`                 ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pairs:

SUM(cons(s(0), cons(s(0), x'''''')), cons(m'', y''')) -> SUM(cons(0, cons(s(0), x'''''')), cons(s(m''), y'''))
SUM(cons(s(0), cons(s(s(n'''''')), x'''''')), cons(m'', y''')) -> SUM(cons(0, cons(s(s(n'''''')), x'''''')), cons(s(m''), y'''))
SUM(cons(0, cons(s(0), x'''')), cons(m''', y'''')) -> SUM(cons(s(0), x''''), cons(m''', y''''))
SUM(cons(s(s(s(n''''))), x''''), cons(m'''', y'''')) -> SUM(cons(s(s(n'''')), x''''), cons(s(m''''), y''''))
SUM(cons(0, cons(s(s(n'''')), x'''')), cons(m'''', y'''')) -> SUM(cons(s(s(n'''')), x''''), cons(m'''', y''''))
SUM(cons(0, cons(0, x'')), y'') -> SUM(cons(0, x''), y'')
SUM(cons(s(0), cons(0, x'''')), cons(m'', y'''')) -> SUM(cons(0, cons(0, x'''')), cons(s(m''), y''''))
SUM(cons(s(s(0)), x''''), cons(m'''', y'''')) -> SUM(cons(s(0), x''''), cons(s(m''''), y''''))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

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

SUM(cons(0, cons(0, x'')), y'') -> SUM(cons(0, x''), y'')
three new Dependency Pairs are created:

SUM(cons(0, cons(0, cons(0, x''''))), y'''') -> SUM(cons(0, cons(0, x'''')), y'''')
SUM(cons(0, cons(0, cons(s(s(n'''''')), x''''''))), cons(m'''''', y'''''')) -> SUM(cons(0, cons(s(s(n'''''')), x'''''')), cons(m'''''', y''''''))
SUM(cons(0, cons(0, cons(s(0), x''''''))), cons(m''''', y'''''')) -> SUM(cons(0, cons(s(0), x'''''')), cons(m''''', y''''''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 7`
`                 ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pairs:

SUM(cons(0, cons(0, cons(s(0), x''''''))), cons(m''''', y'''''')) -> SUM(cons(0, cons(s(0), x'''''')), cons(m''''', y''''''))
SUM(cons(s(0), cons(s(s(n'''''')), x'''''')), cons(m'', y''')) -> SUM(cons(0, cons(s(s(n'''''')), x'''''')), cons(s(m''), y'''))
SUM(cons(s(s(0)), x''''), cons(m'''', y'''')) -> SUM(cons(s(0), x''''), cons(s(m''''), y''''))
SUM(cons(s(s(s(n''''))), x''''), cons(m'''', y'''')) -> SUM(cons(s(s(n'''')), x''''), cons(s(m''''), y''''))
SUM(cons(0, cons(s(s(n'''')), x'''')), cons(m'''', y'''')) -> SUM(cons(s(s(n'''')), x''''), cons(m'''', y''''))
SUM(cons(0, cons(0, cons(s(s(n'''''')), x''''''))), cons(m'''''', y'''''')) -> SUM(cons(0, cons(s(s(n'''''')), x'''''')), cons(m'''''', y''''''))
SUM(cons(0, cons(0, cons(0, x''''))), y'''') -> SUM(cons(0, cons(0, x'''')), y'''')
SUM(cons(s(0), cons(0, x'''')), cons(m'', y'''')) -> SUM(cons(0, cons(0, x'''')), cons(s(m''), y''''))
SUM(cons(0, cons(s(0), x'''')), cons(m''', y'''')) -> SUM(cons(s(0), x''''), cons(m''', y''''))
SUM(cons(s(0), cons(s(0), x'''''')), cons(m'', y''')) -> SUM(cons(0, cons(s(0), x'''''')), cons(s(m''), y'''))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

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

SUM(cons(0, cons(s(s(n'''')), x'''')), cons(m'''', y'''')) -> SUM(cons(s(s(n'''')), x''''), cons(m'''', y''''))
two new Dependency Pairs are created:

SUM(cons(0, cons(s(s(s(n''''''))), x'''''')), cons(m'''''', y'''''')) -> SUM(cons(s(s(s(n''''''))), x''''''), cons(m'''''', y''''''))
SUM(cons(0, cons(s(s(0)), x'''''')), cons(m'''''', y'''''')) -> SUM(cons(s(s(0)), x''''''), cons(m'''''', y''''''))

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

SUM(cons(s(0), cons(s(0), x'''''')), cons(m'', y''')) -> SUM(cons(0, cons(s(0), x'''''')), cons(s(m''), y'''))
SUM(cons(0, cons(s(s(0)), x'''''')), cons(m'''''', y'''''')) -> SUM(cons(s(s(0)), x''''''), cons(m'''''', y''''''))
SUM(cons(s(0), cons(s(s(n'''''')), x'''''')), cons(m'', y''')) -> SUM(cons(0, cons(s(s(n'''''')), x'''''')), cons(s(m''), y'''))
SUM(cons(s(s(0)), x''''), cons(m'''', y'''')) -> SUM(cons(s(0), x''''), cons(s(m''''), y''''))
SUM(cons(s(s(s(n''''))), x''''), cons(m'''', y'''')) -> SUM(cons(s(s(n'''')), x''''), cons(s(m''''), y''''))
SUM(cons(0, cons(s(s(s(n''''''))), x'''''')), cons(m'''''', y'''''')) -> SUM(cons(s(s(s(n''''''))), x''''''), cons(m'''''', y''''''))
SUM(cons(0, cons(0, cons(s(s(n'''''')), x''''''))), cons(m'''''', y'''''')) -> SUM(cons(0, cons(s(s(n'''''')), x'''''')), cons(m'''''', y''''''))
SUM(cons(0, cons(0, cons(0, x''''))), y'''') -> SUM(cons(0, cons(0, x'''')), y'''')
SUM(cons(s(0), cons(0, x'''')), cons(m'', y'''')) -> SUM(cons(0, cons(0, x'''')), cons(s(m''), y''''))
SUM(cons(0, cons(s(0), x'''')), cons(m''', y'''')) -> SUM(cons(s(0), x''''), cons(m''', y''''))
SUM(cons(0, cons(0, cons(s(0), x''''''))), cons(m''''', y'''''')) -> SUM(cons(0, cons(s(0), x'''''')), cons(m''''', y''''''))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

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

SUM(cons(0, cons(s(0), x'''')), cons(m''', y'''')) -> SUM(cons(s(0), x''''), cons(m''', y''''))
three new Dependency Pairs are created:

SUM(cons(0, cons(s(0), cons(0, x''''''))), cons(m''''', y'''''')) -> SUM(cons(s(0), cons(0, x'''''')), cons(m''''', y''''''))
SUM(cons(0, cons(s(0), cons(s(s(n'''''''')), x''''''''))), cons(m''''', y'''''')) -> SUM(cons(s(0), cons(s(s(n'''''''')), x'''''''')), cons(m''''', y''''''))
SUM(cons(0, cons(s(0), cons(s(0), x''''''''))), cons(m''''', y'''''')) -> SUM(cons(s(0), cons(s(0), x'''''''')), cons(m''''', y''''''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 9`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pairs:

SUM(cons(0, cons(s(0), cons(s(0), x''''''''))), cons(m''''', y'''''')) -> SUM(cons(s(0), cons(s(0), x'''''''')), cons(m''''', y''''''))
SUM(cons(0, cons(s(0), cons(s(s(n'''''''')), x''''''''))), cons(m''''', y'''''')) -> SUM(cons(s(0), cons(s(s(n'''''''')), x'''''''')), cons(m''''', y''''''))
SUM(cons(0, cons(0, cons(s(0), x''''''))), cons(m''''', y'''''')) -> SUM(cons(0, cons(s(0), x'''''')), cons(m''''', y''''''))
SUM(cons(0, cons(s(s(0)), x'''''')), cons(m'''''', y'''''')) -> SUM(cons(s(s(0)), x''''''), cons(m'''''', y''''''))
SUM(cons(s(0), cons(s(s(n'''''')), x'''''')), cons(m'', y''')) -> SUM(cons(0, cons(s(s(n'''''')), x'''''')), cons(s(m''), y'''))
SUM(cons(s(s(0)), x''''), cons(m'''', y'''')) -> SUM(cons(s(0), x''''), cons(s(m''''), y''''))
SUM(cons(s(s(s(n''''))), x''''), cons(m'''', y'''')) -> SUM(cons(s(s(n'''')), x''''), cons(s(m''''), y''''))
SUM(cons(0, cons(s(s(s(n''''''))), x'''''')), cons(m'''''', y'''''')) -> SUM(cons(s(s(s(n''''''))), x''''''), cons(m'''''', y''''''))
SUM(cons(0, cons(0, cons(s(s(n'''''')), x''''''))), cons(m'''''', y'''''')) -> SUM(cons(0, cons(s(s(n'''''')), x'''''')), cons(m'''''', y''''''))
SUM(cons(0, cons(0, cons(0, x''''))), y'''') -> SUM(cons(0, cons(0, x'''')), y'''')
SUM(cons(s(0), cons(0, x'''')), cons(m'', y'''')) -> SUM(cons(0, cons(0, x'''')), cons(s(m''), y''''))
SUM(cons(0, cons(s(0), cons(0, x''''''))), cons(m''''', y'''''')) -> SUM(cons(s(0), cons(0, x'''''')), cons(m''''', y''''''))
SUM(cons(s(0), cons(s(0), x'''''')), cons(m'', y''')) -> SUM(cons(0, cons(s(0), x'''''')), cons(s(m''), y'''))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

The following dependency pairs can be strictly oriented:

SUM(cons(0, cons(s(0), cons(s(0), x''''''''))), cons(m''''', y'''''')) -> SUM(cons(s(0), cons(s(0), x'''''''')), cons(m''''', y''''''))
SUM(cons(0, cons(s(0), cons(s(s(n'''''''')), x''''''''))), cons(m''''', y'''''')) -> SUM(cons(s(0), cons(s(s(n'''''''')), x'''''''')), cons(m''''', y''''''))
SUM(cons(0, cons(0, cons(s(0), x''''''))), cons(m''''', y'''''')) -> SUM(cons(0, cons(s(0), x'''''')), cons(m''''', y''''''))
SUM(cons(0, cons(s(s(0)), x'''''')), cons(m'''''', y'''''')) -> SUM(cons(s(s(0)), x''''''), cons(m'''''', y''''''))
SUM(cons(0, cons(s(s(s(n''''''))), x'''''')), cons(m'''''', y'''''')) -> SUM(cons(s(s(s(n''''''))), x''''''), cons(m'''''', y''''''))
SUM(cons(0, cons(0, cons(s(s(n'''''')), x''''''))), cons(m'''''', y'''''')) -> SUM(cons(0, cons(s(s(n'''''')), x'''''')), cons(m'''''', y''''''))
SUM(cons(0, cons(0, cons(0, x''''))), y'''') -> SUM(cons(0, cons(0, x'''')), y'''')
SUM(cons(0, cons(s(0), cons(0, x''''''))), cons(m''''', y'''''')) -> SUM(cons(s(0), cons(0, x'''''')), cons(m''''', 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(SUM(x1, x2)) =  x1 POL(0) =  0 POL(cons(x1, x2)) =  1 + x2 POL(s(x1)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 10`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pairs:

SUM(cons(s(0), cons(s(s(n'''''')), x'''''')), cons(m'', y''')) -> SUM(cons(0, cons(s(s(n'''''')), x'''''')), cons(s(m''), y'''))
SUM(cons(s(s(0)), x''''), cons(m'''', y'''')) -> SUM(cons(s(0), x''''), cons(s(m''''), y''''))
SUM(cons(s(s(s(n''''))), x''''), cons(m'''', y'''')) -> SUM(cons(s(s(n'''')), x''''), cons(s(m''''), y''''))
SUM(cons(s(0), cons(0, x'''')), cons(m'', y'''')) -> SUM(cons(0, cons(0, x'''')), cons(s(m''), y''''))
SUM(cons(s(0), cons(s(0), x'''''')), cons(m'', y''')) -> SUM(cons(0, cons(s(0), x'''''')), cons(s(m''), y'''))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 11`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

SUM(cons(s(s(s(n''''))), x''''), cons(m'''', y'''')) -> SUM(cons(s(s(n'''')), x''''), cons(s(m''''), y''''))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

The following dependency pair can be strictly oriented:

SUM(cons(s(s(s(n''''))), x''''), cons(m'''', y'''')) -> SUM(cons(s(s(n'''')), x''''), cons(s(m''''), 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(SUM(x1, x2)) =  1 + x1 POL(cons(x1, x2)) =  1 + x1 + x2 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 12`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

WEIGHT(cons(n, cons(m, x))) -> WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

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

WEIGHT(cons(n, cons(m, x))) -> WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))
two new Dependency Pairs are created:

WEIGHT(cons(s(n''), cons(m'', x''))) -> WEIGHT(sum(cons(n'', cons(m'', x'')), cons(s(0), x'')))
WEIGHT(cons(0, cons(m', x''))) -> WEIGHT(sum(cons(m', x''), cons(0, x'')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 13`
`             ↳Narrowing Transformation`

Dependency Pairs:

WEIGHT(cons(0, cons(m', x''))) -> WEIGHT(sum(cons(m', x''), cons(0, x'')))
WEIGHT(cons(s(n''), cons(m'', x''))) -> WEIGHT(sum(cons(n'', cons(m'', x'')), cons(s(0), x'')))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

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

WEIGHT(cons(s(n''), cons(m'', x''))) -> WEIGHT(sum(cons(n'', cons(m'', x'')), cons(s(0), x'')))
two new Dependency Pairs are created:

WEIGHT(cons(s(s(n')), cons(m''', x'''))) -> WEIGHT(sum(cons(n', cons(m''', x''')), cons(s(s(0)), x''')))
WEIGHT(cons(s(0), cons(m''', x'''))) -> WEIGHT(sum(cons(m''', x'''), cons(s(0), x''')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 13`
`             ↳Nar`
`             ...`
`               →DP Problem 14`
`                 ↳Narrowing Transformation`

Dependency Pairs:

WEIGHT(cons(s(0), cons(m''', x'''))) -> WEIGHT(sum(cons(m''', x'''), cons(s(0), x''')))
WEIGHT(cons(s(s(n')), cons(m''', x'''))) -> WEIGHT(sum(cons(n', cons(m''', x''')), cons(s(s(0)), x''')))
WEIGHT(cons(0, cons(m', x''))) -> WEIGHT(sum(cons(m', x''), cons(0, x'')))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

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

WEIGHT(cons(0, cons(m', x''))) -> WEIGHT(sum(cons(m', x''), cons(0, x'')))
two new Dependency Pairs are created:

WEIGHT(cons(0, cons(s(n'), x'''))) -> WEIGHT(sum(cons(n', x'''), cons(s(0), x''')))
WEIGHT(cons(0, cons(0, x'''))) -> WEIGHT(sum(x''', cons(0, x''')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 13`
`             ↳Nar`
`             ...`
`               →DP Problem 15`
`                 ↳Polynomial Ordering`

Dependency Pairs:

WEIGHT(cons(0, cons(0, x'''))) -> WEIGHT(sum(x''', cons(0, x''')))
WEIGHT(cons(0, cons(s(n'), x'''))) -> WEIGHT(sum(cons(n', x'''), cons(s(0), x''')))
WEIGHT(cons(s(s(n')), cons(m''', x'''))) -> WEIGHT(sum(cons(n', cons(m''', x''')), cons(s(s(0)), x''')))
WEIGHT(cons(s(0), cons(m''', x'''))) -> WEIGHT(sum(cons(m''', x'''), cons(s(0), x''')))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

The following dependency pair can be strictly oriented:

WEIGHT(cons(0, cons(s(n'), x'''))) -> WEIGHT(sum(cons(n', x'''), cons(s(0), x''')))

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

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  1 POL(cons(x1, x2)) =  x1 POL(WEIGHT(x1)) =  x1 POL(nil) =  1 POL(sum(x1, x2)) =  x2 POL(s(x1)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 13`
`             ↳Nar`
`             ...`
`               →DP Problem 16`
`                 ↳Polynomial Ordering`

Dependency Pairs:

WEIGHT(cons(0, cons(0, x'''))) -> WEIGHT(sum(x''', cons(0, x''')))
WEIGHT(cons(s(s(n')), cons(m''', x'''))) -> WEIGHT(sum(cons(n', cons(m''', x''')), cons(s(s(0)), x''')))
WEIGHT(cons(s(0), cons(m''', x'''))) -> WEIGHT(sum(cons(m''', x'''), cons(s(0), x''')))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

The following dependency pair can be strictly oriented:

WEIGHT(cons(0, cons(0, x'''))) -> WEIGHT(sum(x''', cons(0, x''')))

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

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  1 POL(cons(x1, x2)) =  x1 + x2 POL(WEIGHT(x1)) =  x1 POL(nil) =  1 POL(sum(x1, x2)) =  x2 POL(s(x1)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 13`
`             ↳Nar`
`             ...`
`               →DP Problem 17`
`                 ↳Polynomial Ordering`

Dependency Pairs:

WEIGHT(cons(s(s(n')), cons(m''', x'''))) -> WEIGHT(sum(cons(n', cons(m''', x''')), cons(s(s(0)), x''')))
WEIGHT(cons(s(0), cons(m''', x'''))) -> WEIGHT(sum(cons(m''', x'''), cons(s(0), x''')))

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

The following dependency pairs can be strictly oriented:

WEIGHT(cons(s(s(n')), cons(m''', x'''))) -> WEIGHT(sum(cons(n', cons(m''', x''')), cons(s(s(0)), x''')))
WEIGHT(cons(s(0), cons(m''', x'''))) -> WEIGHT(sum(cons(m''', x'''), cons(s(0), x''')))

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

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(cons(x1, x2)) =  1 + x2 POL(WEIGHT(x1)) =  1 + x1 POL(nil) =  0 POL(sum(x1, x2)) =  x2 POL(s(x1)) =  0

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 13`
`             ↳Nar`
`             ...`
`               →DP Problem 18`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:02 minutes