Termination Proof using AProVE

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.

     ↳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.

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳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:

     ↳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 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.

     ↳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.

     ↳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 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.

     ↳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.

     ↳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:

     ↳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:

     ↳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:

     ↳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 pairs can be strictly oriented:

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''')))

Additionally, the following usable rules for innermost 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.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 13

             ↳Nar

...

               →DP Problem 16

                 ↳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 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.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 13

             ↳Nar

...

               →DP Problem 17

                 ↳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:
0:01 minutes

POL(SUM(x₁, x₂))	= x₁
POL(0)	= 0
POL(cons(x₁, x₂))	= 1 + x₂
POL(s(x₁))	= 0

POL(SUM(x₁, x₂))	= 1 + x₁
POL(cons(x₁, x₂))	= 1 + x₁ + x₂
POL(s(x₁))	= 1 + x₁

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