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:
g1(X) -> u3(h1(X), h1(X), X)
u3(d, c1(Y), X) -> k1(Y)
h1(d) -> c1(a)
h1(d) -> c1(b)
f3(k1(a), k1(b), X) -> f3(X, X, X)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g1(X) -> u3(h1(X), h1(X), X)
u3(d, c1(Y), X) -> k1(Y)
h1(d) -> c1(a)
h1(d) -> c1(b)
f3(k1(a), k1(b), X) -> f3(X, X, 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:
G1(X) -> H1(X)
F3(k1(a), k1(b), X) -> F3(X, X, X)
G1(X) -> U3(h1(X), h1(X), X)
The TRS R consists of the following rules:
g1(X) -> u3(h1(X), h1(X), X)
u3(d, c1(Y), X) -> k1(Y)
h1(d) -> c1(a)
h1(d) -> c1(b)
f3(k1(a), k1(b), X) -> f3(X, X, X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
G1(X) -> H1(X)
F3(k1(a), k1(b), X) -> F3(X, X, X)
G1(X) -> U3(h1(X), h1(X), X)
The TRS R consists of the following rules:
g1(X) -> u3(h1(X), h1(X), X)
u3(d, c1(Y), X) -> k1(Y)
h1(d) -> c1(a)
h1(d) -> c1(b)
f3(k1(a), k1(b), X) -> f3(X, X, X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F3(k1(a), k1(b), X) -> F3(X, X, X)
The TRS R consists of the following rules:
g1(X) -> u3(h1(X), h1(X), X)
u3(d, c1(Y), X) -> k1(Y)
h1(d) -> c1(a)
h1(d) -> c1(b)
f3(k1(a), k1(b), X) -> f3(X, X, X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.