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) -> h1(X)
c -> d
h1(d) -> g1(c)
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
g1(X) -> h1(X)
c -> d
h1(d) -> g1(c)
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:
g1(X) -> h1(X)
c -> d
h1(d) -> g1(c)
The set Q consists of the following terms:
g1(x0)
c
h1(d)
Q DP problem:
The TRS P consists of the following rules:
G1(X) -> H1(X)
H1(d) -> G1(c)
H1(d) -> C
The TRS R consists of the following rules:
g1(X) -> h1(X)
c -> d
h1(d) -> g1(c)
The set Q consists of the following terms:
g1(x0)
c
h1(d)
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(X) -> H1(X)
H1(d) -> G1(c)
H1(d) -> C
The TRS R consists of the following rules:
g1(X) -> h1(X)
c -> d
h1(d) -> g1(c)
The set Q consists of the following terms:
g1(x0)
c
h1(d)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G1(X) -> H1(X)
H1(d) -> G1(c)
The TRS R consists of the following rules:
g1(X) -> h1(X)
c -> d
h1(d) -> g1(c)
The set Q consists of the following terms:
g1(x0)
c
h1(d)
We have to consider all minimal (P,Q,R)-chains.