Termination Proof using AProVE

Term Rewriting System R:
[xs, ys, x, y, p]
app(app(app(if, true), xs), ys) -> xs
app(app(app(if, false), xs), ys) -> ys
app(app(sub, x), 0) -> x
app(app(sub, app(s, x)), app(s, y)) -> app(app(sub, x), y)
app(app(gtr, 0), y) -> false
app(app(gtr, app(s, x)), 0) -> true
app(app(gtr, app(s, x)), app(s, y)) -> app(app(gtr, x), y)
app(app(d, x), 0) -> true
app(app(d, app(s, x)), app(s, y)) -> app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) -> 0
app(len, app(app(cons, x), xs)) -> app(s, app(len, xs))
app(app(filter, p), nil) -> nil
app(app(filter, p), app(app(cons, x), xs)) -> app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(app(sub, app(s, x)), app(s, y)) -> APP(app(sub, x), y)
APP(app(sub, app(s, x)), app(s, y)) -> APP(sub, x)
APP(app(gtr, app(s, x)), app(s, y)) -> APP(app(gtr, x), y)
APP(app(gtr, app(s, x)), app(s, y)) -> APP(gtr, x)
APP(app(d, app(s, x)), app(s, y)) -> APP(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
APP(app(d, app(s, x)), app(s, y)) -> APP(app(if, app(app(gtr, x), y)), false)
APP(app(d, app(s, x)), app(s, y)) -> APP(if, app(app(gtr, x), y))
APP(app(d, app(s, x)), app(s, y)) -> APP(app(gtr, x), y)
APP(app(d, app(s, x)), app(s, y)) -> APP(gtr, x)
APP(app(d, app(s, x)), app(s, y)) -> APP(app(d, app(s, x)), app(app(sub, y), x))
APP(app(d, app(s, x)), app(s, y)) -> APP(app(sub, y), x)
APP(app(d, app(s, x)), app(s, y)) -> APP(sub, y)
APP(len, app(app(cons, x), xs)) -> APP(s, app(len, xs))
APP(len, app(app(cons, x), xs)) -> APP(len, xs)
APP(app(filter, p), app(app(cons, x), xs)) -> APP(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))
APP(app(filter, p), app(app(cons, x), xs)) -> APP(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs)))
APP(app(filter, p), app(app(cons, x), xs)) -> APP(if, app(p, x))
APP(app(filter, p), app(app(cons, x), xs)) -> APP(p, x)
APP(app(filter, p), app(app(cons, x), xs)) -> APP(app(cons, x), app(app(filter, p), xs))
APP(app(filter, p), app(app(cons, x), xs)) -> APP(app(filter, p), xs)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Remaining

Dependency Pair:

APP(len, app(app(cons, x), xs)) -> APP(len, xs)

Rules:

app(app(app(if, true), xs), ys) -> xs
app(app(app(if, false), xs), ys) -> ys
app(app(sub, x), 0) -> x
app(app(sub, app(s, x)), app(s, y)) -> app(app(sub, x), y)
app(app(gtr, 0), y) -> false
app(app(gtr, app(s, x)), 0) -> true
app(app(gtr, app(s, x)), app(s, y)) -> app(app(gtr, x), y)
app(app(d, x), 0) -> true
app(app(d, app(s, x)), app(s, y)) -> app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) -> 0
app(len, app(app(cons, x), xs)) -> app(s, app(len, xs))
app(app(filter, p), nil) -> nil
app(app(filter, p), app(app(cons, x), xs)) -> app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

Strategy:

innermost

The following dependency pair can be strictly oriented:

APP(len, app(app(cons, x), xs)) -> APP(len, xs)

There are no usable rules for innermost 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

         ↳Remaining

Dependency Pair:

Rules:

app(app(app(if, true), xs), ys) -> xs
app(app(app(if, false), xs), ys) -> ys
app(app(sub, x), 0) -> x
app(app(sub, app(s, x)), app(s, y)) -> app(app(sub, x), y)
app(app(gtr, 0), y) -> false
app(app(gtr, app(s, x)), 0) -> true
app(app(gtr, app(s, x)), app(s, y)) -> app(app(gtr, x), y)
app(app(d, x), 0) -> true
app(app(d, app(s, x)), app(s, y)) -> app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) -> 0
app(len, app(app(cons, x), xs)) -> app(s, app(len, xs))
app(app(filter, p), nil) -> nil
app(app(filter, p), app(app(cons, x), xs)) -> app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

APP(app(filter, p), app(app(cons, x), xs)) -> APP(app(filter, p), xs)
APP(app(filter, p), app(app(cons, x), xs)) -> APP(p, x)
APP(app(filter, p), app(app(cons, x), xs)) -> APP(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs)))
APP(app(filter, p), app(app(cons, x), xs)) -> APP(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))
APP(app(d, app(s, x)), app(s, y)) -> APP(app(sub, y), x)
APP(app(d, app(s, x)), app(s, y)) -> APP(app(d, app(s, x)), app(app(sub, y), x))
APP(app(d, app(s, x)), app(s, y)) -> APP(app(gtr, x), y)
APP(app(d, app(s, x)), app(s, y)) -> APP(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
APP(app(gtr, app(s, x)), app(s, y)) -> APP(app(gtr, x), y)
APP(app(sub, app(s, x)), app(s, y)) -> APP(app(sub, x), y)

Rules:

app(app(app(if, true), xs), ys) -> xs
app(app(app(if, false), xs), ys) -> ys
app(app(sub, x), 0) -> x
app(app(sub, app(s, x)), app(s, y)) -> app(app(sub, x), y)
app(app(gtr, 0), y) -> false
app(app(gtr, app(s, x)), 0) -> true
app(app(gtr, app(s, x)), app(s, y)) -> app(app(gtr, x), y)
app(app(d, x), 0) -> true
app(app(d, app(s, x)), app(s, y)) -> app(app(app(if, app(app(gtr, x), y)), false), app(app(d, app(s, x)), app(app(sub, y), x)))
app(len, nil) -> 0
app(len, app(app(cons, x), xs)) -> app(s, app(len, xs))
app(app(filter, p), nil) -> nil
app(app(filter, p), app(app(cons, x), xs)) -> app(app(app(if, app(p, x)), app(app(cons, x), app(app(filter, p), xs))), app(app(filter, p), xs))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:11 minutes