Termination Proof using AProVE

Term Rewriting System R:
[y, z, x]
app(f, app(app(cons, nil), y)) -> y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) -> app(f, z)
app(app(app(copy, app(s, x)), y), z) -> app(app(app(copy, x), y), app(app(cons, app(f, y)), z))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> APP(app(app(copy, n), y), z)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> APP(app(copy, n), y)
APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> APP(copy, n)
APP(app(app(copy, 0), y), z) -> APP(f, z)
APP(app(app(copy, app(s, x)), y), z) -> APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))
APP(app(app(copy, app(s, x)), y), z) -> APP(app(copy, x), y)
APP(app(app(copy, app(s, x)), y), z) -> APP(copy, x)
APP(app(app(copy, app(s, x)), y), z) -> APP(app(cons, app(f, y)), z)
APP(app(app(copy, app(s, x)), y), z) -> APP(cons, app(f, y))
APP(app(app(copy, app(s, x)), y), z) -> APP(f, y)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

APP(app(app(copy, app(s, x)), y), z) -> APP(app(cons, app(f, y)), z)
APP(app(app(copy, app(s, x)), y), z) -> APP(app(copy, x), y)
APP(app(app(copy, app(s, x)), y), z) -> APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))

Rules:

app(f, app(app(cons, nil), y)) -> y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) -> app(f, z)
app(app(app(copy, app(s, x)), y), z) -> app(app(app(copy, x), y), app(app(cons, app(f, y)), z))

Strategy:

innermost

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

APP(app(app(copy, app(s, x)), y), z) -> APP(app(copy, x), y)

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

APP(app(app(copy, app(s, x)), y), z) -> APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))
APP(app(app(copy, app(s, x)), y), z) -> APP(app(cons, app(f, y)), z)

Rules:

app(f, app(app(cons, nil), y)) -> y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) -> app(f, z)
app(app(app(copy, app(s, x)), y), z) -> app(app(app(copy, x), y), app(app(cons, app(f, y)), z))

Strategy:

innermost

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

APP(app(app(copy, app(s, x)), y), z) -> APP(app(cons, app(f, y)), z)

one new Dependency Pair is created:

APP(app(app(copy, app(s, x)), app(app(cons, nil), y'')), z) -> APP(app(cons, y''), z)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Narrowing Transformation

Dependency Pairs:

APP(app(app(copy, app(s, x)), app(app(cons, nil), y'')), z) -> APP(app(cons, y''), z)
APP(app(app(copy, app(s, x)), y), z) -> APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))

Rules:

app(f, app(app(cons, nil), y)) -> y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) -> app(f, z)
app(app(app(copy, app(s, x)), y), z) -> app(app(app(copy, x), y), app(app(cons, app(f, y)), z))

Strategy:

innermost

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

APP(app(app(copy, app(s, x)), app(app(cons, nil), y'')), z) -> APP(app(cons, y''), z)

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Argument Filtering and Ordering

Dependency Pair:

APP(app(app(copy, app(s, x)), y), z) -> APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))

Rules:

app(f, app(app(cons, nil), y)) -> y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) -> app(f, z)
app(app(app(copy, app(s, x)), y), z) -> app(app(app(copy, x), y), app(app(cons, app(f, y)), z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

APP(app(app(copy, app(s, x)), y), z) -> APP(app(app(copy, x), y), app(app(cons, app(f, y)), z))

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

app(f, app(app(cons, nil), y)) -> y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) -> app(f, z)
app(app(app(copy, app(s, x)), y), z) -> app(app(app(copy, x), y), app(app(cons, app(f, y)), z))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

APP > {cons, f, app, n} > copy

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

APP(x₁, x₂) -> APP(x₁, x₂)
app(x₁, x₂) -> app(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Dependency Graph

Dependency Pair:

Rules:

app(f, app(app(cons, nil), y)) -> y
app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> app(app(app(copy, n), y), z)
app(app(app(copy, 0), y), z) -> app(f, z)
app(app(app(copy, app(s, x)), y), z) -> app(app(app(copy, x), y), app(app(cons, app(f, y)), z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:21 minutes