Termination Proof using AProVE

Term Rewriting System R:
[x, y]
app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(double, 0) -> 0
app(double, app(s, x)) -> app(s, app(s, app(double, x)))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, y)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(minus, app(s, x)), app(s, y)) -> APP(minus, x)
APP(double, app(s, x)) -> APP(s, app(s, app(double, x)))
APP(double, app(s, x)) -> APP(s, app(double, x))
APP(double, app(s, x)) -> APP(double, x)
APP(app(plus, app(s, x)), y) -> APP(s, app(app(plus, x), y))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) -> APP(plus, x)
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), app(s, y))
APP(app(plus, app(s, x)), y) -> APP(s, y)
APP(app(plus, app(s, x)), y) -> APP(s, app(app(plus, app(app(minus, x), y)), app(double, y)))
APP(app(plus, app(s, x)), y) -> APP(app(plus, app(app(minus, x), y)), app(double, y))
APP(app(plus, app(s, x)), y) -> APP(plus, app(app(minus, x), y))
APP(app(plus, app(s, x)), y) -> APP(app(minus, x), y)
APP(app(plus, app(s, x)), y) -> APP(minus, x)
APP(app(plus, app(s, x)), y) -> APP(double, y)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Nar

Dependency Pair:

APP(double, app(s, x)) -> APP(double, x)

Rules:

app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(double, 0) -> 0
app(double, app(s, x)) -> app(s, app(s, app(double, x)))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

APP(double, app(s, x)) -> APP(double, x)

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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

         ↳AFS

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

Rules:

app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(double, 0) -> 0
app(double, app(s, x)) -> app(s, app(s, app(double, x)))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, y)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Narrowing Transformation

Dependency Pairs:

APP(app(plus, app(s, x)), y) -> APP(app(minus, x), y)
APP(app(plus, app(s, x)), y) -> APP(app(plus, app(app(minus, x), y)), app(double, y))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), app(s, y))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)

Rules:

app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(double, 0) -> 0
app(double, app(s, x)) -> app(s, app(s, app(double, x)))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, y)))

Strategy:

innermost

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

APP(app(plus, app(s, x)), y) -> APP(app(plus, app(app(minus, x), y)), app(double, y))

four new Dependency Pairs are created:

APP(app(plus, app(s, x'')), 0) -> APP(app(plus, x''), app(double, 0))
APP(app(plus, app(s, app(s, x''))), app(s, y'')) -> APP(app(plus, app(app(minus, x''), y'')), app(double, app(s, y'')))
APP(app(plus, app(s, x)), 0) -> APP(app(plus, app(app(minus, x), 0)), 0)
APP(app(plus, app(s, x)), app(s, x'')) -> APP(app(plus, app(app(minus, x), app(s, x''))), app(s, app(s, app(double, x''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

APP(app(plus, app(s, x)), app(s, x'')) -> APP(app(plus, app(app(minus, x), app(s, x''))), app(s, app(s, app(double, x''))))
APP(app(plus, app(s, x)), 0) -> APP(app(plus, app(app(minus, x), 0)), 0)
APP(app(plus, app(s, app(s, x''))), app(s, y'')) -> APP(app(plus, app(app(minus, x''), y'')), app(double, app(s, y'')))
APP(app(plus, app(s, x'')), 0) -> APP(app(plus, x''), app(double, 0))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), app(s, y))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(plus, app(s, x)), y) -> APP(app(minus, x), y)

Rules:

app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(double, 0) -> 0
app(double, app(s, x)) -> app(s, app(s, app(double, x)))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, app(s, x)), y) -> app(app(plus, x), app(s, y))
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, app(app(minus, x), y)), app(double, y)))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:04 minutes