Termination Proof using AProVE

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

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(app(quot, app(s, x)), app(s, y)) -> APP(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(quot, app(s, x)), app(s, y)) -> APP(quot, app(app(minus, x), y))
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) -> APP(minus, 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(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) -> APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) -> APP(plus, app(app(minus, y), app(s, app(s, z))))
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) -> APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) -> APP(plus, app(app(plus, y), app(s, app(s, z))))

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) -> APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) -> APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(quot, app(app(minus, x), y)), app(s, 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(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) -> app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) -> app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))

Strategy:

innermost

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

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

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(quot, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) -> APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))

Rules:

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

Strategy:

innermost

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

APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) -> APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))

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 3

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(quot, 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(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) -> app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) -> app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))

Strategy:

innermost

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