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:
f1(X) -> if3(X, c, f1(true))
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(X) -> if3(X, c, f1(true))
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
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:
f1(X) -> if3(X, c, f1(true))
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
The set Q consists of the following terms:
f1(x0)
if3(true, x0, x1)
if3(false, 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:
F1(X) -> F1(true)
F1(X) -> IF3(X, c, f1(true))
The TRS R consists of the following rules:
f1(X) -> if3(X, c, f1(true))
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
The set Q consists of the following terms:
f1(x0)
if3(true, x0, x1)
if3(false, 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:
F1(X) -> F1(true)
F1(X) -> IF3(X, c, f1(true))
The TRS R consists of the following rules:
f1(X) -> if3(X, c, f1(true))
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
The set Q consists of the following terms:
f1(x0)
if3(true, x0, x1)
if3(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] 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:
F1(X) -> F1(true)
The TRS R consists of the following rules:
f1(X) -> if3(X, c, f1(true))
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
The set Q consists of the following terms:
f1(x0)
if3(true, x0, x1)
if3(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.