Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(x, y) -> x
g1(a) -> h3(a, b, a)
i1(x) -> f2(x, x)
h3(x, x, y) -> g1(x)
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
f2(x, y) -> x
g1(a) -> h3(a, b, a)
i1(x) -> f2(x, x)
h3(x, x, y) -> g1(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:
f2(x, y) -> x
g1(a) -> h3(a, b, a)
i1(x) -> f2(x, x)
h3(x, x, y) -> g1(x)
The set Q consists of the following terms:
f2(x0, x1)
g1(a)
i1(x0)
h3(x0, x0, x1)
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
G1(a) -> H3(a, b, a)
H3(x, x, y) -> G1(x)
I1(x) -> F2(x, x)
The TRS R consists of the following rules:
f2(x, y) -> x
g1(a) -> h3(a, b, a)
i1(x) -> f2(x, x)
h3(x, x, y) -> g1(x)
The set Q consists of the following terms:
f2(x0, x1)
g1(a)
i1(x0)
h3(x0, x0, x1)
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:
G1(a) -> H3(a, b, a)
H3(x, x, y) -> G1(x)
I1(x) -> F2(x, x)
The TRS R consists of the following rules:
f2(x, y) -> x
g1(a) -> h3(a, b, a)
i1(x) -> f2(x, x)
h3(x, x, y) -> g1(x)
The set Q consists of the following terms:
f2(x0, x1)
g1(a)
i1(x0)
h3(x0, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 3 less nodes.