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(const, x), y) -> x
app2(app2(app2(subst, f), g), x) -> app2(app2(f, x), app2(g, x))
app2(app2(fix, f), x) -> app2(app2(f, app2(fix, f)), x)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
app2(app2(const, x), y) -> x
app2(app2(app2(subst, f), g), x) -> app2(app2(f, x), app2(g, x))
app2(app2(fix, f), x) -> app2(app2(f, app2(fix, f)), x)
Q is empty.
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(fix, f), x) -> APP2(f, app2(fix, f))
APP2(app2(fix, f), x) -> APP2(app2(f, app2(fix, f)), x)
APP2(app2(app2(subst, f), g), x) -> APP2(f, x)
APP2(app2(app2(subst, f), g), x) -> APP2(app2(f, x), app2(g, x))
APP2(app2(app2(subst, f), g), x) -> APP2(g, x)
The TRS R consists of the following rules:
app2(app2(const, x), y) -> x
app2(app2(app2(subst, f), g), x) -> app2(app2(f, x), app2(g, x))
app2(app2(fix, f), x) -> app2(app2(f, app2(fix, f)), x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
APP2(app2(fix, f), x) -> APP2(f, app2(fix, f))
APP2(app2(fix, f), x) -> APP2(app2(f, app2(fix, f)), x)
APP2(app2(app2(subst, f), g), x) -> APP2(f, x)
APP2(app2(app2(subst, f), g), x) -> APP2(app2(f, x), app2(g, x))
APP2(app2(app2(subst, f), g), x) -> APP2(g, x)
The TRS R consists of the following rules:
app2(app2(const, x), y) -> x
app2(app2(app2(subst, f), g), x) -> app2(app2(f, x), app2(g, x))
app2(app2(fix, f), x) -> app2(app2(f, app2(fix, f)), x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.