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

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

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Remaining

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

The following dependency pair can be strictly oriented:

SUM(cons(0, x), y) -> SUM(x, y)

The following rules 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
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(SUM(x₁, x₂)) = 1 + x₁ + x₂

POL(0) = 0

POL(cons(x₁, x₂)) = 1 + x₁ + x₂

POL(nil) = 1

POL(s(x₁)) = x₁

POL(weight(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

SUM(x₁, x₂) -> SUM(x₁, x₂)
cons(x₁, x₂) -> cons(x₁, x₂)
s(x₁) -> s(x₁)
sum(x₁, x₂) -> x₂
weight(x₁) -> weight(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pair:

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

On this DP problem, an 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))

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳Inst

...

               →DP Problem 4

                 ↳Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pair:

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳Inst

...

               →DP Problem 5

                 ↳Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pair:

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳Inst

...

               →DP Problem 6

                 ↳Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pair:

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳Inst

...

               →DP Problem 7

                 ↳Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pair:

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

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

Termination of R could not be shown.
Duration:
0:00 minutes

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