Termination Proof using AProVE

Term Rewriting System R:
[x, l, y]
app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(rev, app(app(cons, x), l)) -> APP(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
APP(rev, app(app(cons, x), l)) -> APP(cons, app(app(rev1, x), l))
APP(rev, app(app(cons, x), l)) -> APP(app(rev1, x), l)
APP(rev, app(app(cons, x), l)) -> APP(rev1, x)
APP(rev, app(app(cons, x), l)) -> APP(app(rev2, x), l)
APP(rev, app(app(cons, x), l)) -> APP(rev2, x)
APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)
APP(app(rev1, x), app(app(cons, y), l)) -> APP(rev1, y)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(cons, x), app(app(rev2, y), l))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(cons, x)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(rev2, y), l)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev2, y)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

Dependency Pairs:

APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(rev2, y), l)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(cons, x), app(app(rev2, y), l))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev2, x), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev1, x), l)
APP(rev, app(app(cons, x), l)) -> APP(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))

Rules:

app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(cons, x), app(app(rev2, y), l))
APP(rev, app(app(cons, x), l)) -> APP(app(rev1, x), l)
APP(rev, app(app(cons, x), l)) -> APP(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))

The following usable rules for innermost can be oriented:

app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))

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

{nil, rev2, rev} > cons
{nil, rev2, rev} > rev1 > 0
{nil, rev2, rev} > rev1 > s

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

APP(x₁, x₂) -> x₁
app(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 2

             ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(rev2, y), l)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)
APP(rev, app(app(cons, x), l)) -> APP(app(rev2, x), l)

Rules:

app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:03 minutes