Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))
Q is empty.
The TRS is non-overlapping. Hence, we can switch to innermost.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))
The set Q consists of the following terms:
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(app2(until, x0), x1), x2)
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(app2(until, p), f), x) -> APP2(p, x)
APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))
APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(until, p), f), app2(f, x))
APP2(app2(app2(until, p), f), x) -> APP2(app2(if, app2(p, x)), x)
APP2(app2(app2(until, p), f), x) -> APP2(if, app2(p, x))
APP2(app2(app2(until, p), f), x) -> APP2(f, x)
The TRS R consists of the following rules:
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))
The set Q consists of the following terms:
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(app2(until, x0), x1), x2)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(app2(until, p), f), x) -> APP2(p, x)
APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))
APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(until, p), f), app2(f, x))
APP2(app2(app2(until, p), f), x) -> APP2(app2(if, app2(p, x)), x)
APP2(app2(app2(until, p), f), x) -> APP2(if, app2(p, x))
APP2(app2(app2(until, p), f), x) -> APP2(f, x)
The TRS R consists of the following rules:
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))
The set Q consists of the following terms:
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(app2(until, x0), x1), x2)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(app2(until, p), f), x) -> APP2(p, x)
APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(until, p), f), app2(f, x))
APP2(app2(app2(until, p), f), x) -> APP2(f, x)
The TRS R consists of the following rules:
app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))
The set Q consists of the following terms:
app2(app2(app2(if, true), x0), x1)
app2(app2(app2(if, false), x0), x1)
app2(app2(app2(until, x0), x1), x2)
We have to consider all minimal (P,Q,R)-chains.