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:
f(x, f(a, f(f(a, a), a))) → f(f(a, x), x)
Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, f(a, f(f(a, a), a))) → f(f(a, x), x)
Q is empty.
The TRS is overlay and locally confluent. By [19] we can switch to innermost.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, f(a, f(f(a, a), a))) → f(f(a, x), x)
The set Q consists of the following terms:
f(x0, f(a, f(f(a, a), a)))
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
F(x, f(a, f(f(a, a), a))) → F(f(a, x), x)
F(x, f(a, f(f(a, a), a))) → F(a, x)
The TRS R consists of the following rules:
f(x, f(a, f(f(a, a), a))) → f(f(a, x), x)
The set Q consists of the following terms:
f(x0, f(a, f(f(a, a), a)))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
F(x, f(a, f(f(a, a), a))) → F(f(a, x), x)
F(x, f(a, f(f(a, a), a))) → F(a, x)
The TRS R consists of the following rules:
f(x, f(a, f(f(a, a), a))) → f(f(a, x), x)
The set Q consists of the following terms:
f(x0, f(a, f(f(a, a), a)))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule F(x, f(a, f(f(a, a), a))) → F(f(a, x), x) at position [] we obtained the following new rules:
F(f(a, f(f(a, a), a)), f(a, f(f(a, a), a))) → F(f(f(a, a), a), f(a, f(f(a, a), a)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(f(a, f(f(a, a), a)), f(a, f(f(a, a), a))) → F(f(f(a, a), a), f(a, f(f(a, a), a)))
F(x, f(a, f(f(a, a), a))) → F(a, x)
The TRS R consists of the following rules:
f(x, f(a, f(f(a, a), a))) → f(f(a, x), x)
The set Q consists of the following terms:
f(x0, f(a, f(f(a, a), a)))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(x, f(a, f(f(a, a), a))) → F(a, x)
The TRS R consists of the following rules:
f(x, f(a, f(f(a, a), a))) → f(f(a, x), x)
The set Q consists of the following terms:
f(x0, f(a, f(f(a, a), a)))
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
F(x, f(a, f(f(a, a), a))) → F(a, x)
R is empty.
The set Q consists of the following terms:
f(x0, f(a, f(f(a, a), a)))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(x, f(a, f(f(a, a), a))) → F(a, x) we obtained the following new rules:
F(a, f(a, f(f(a, a), a))) → F(a, a)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(a, f(a, f(f(a, a), a))) → F(a, a)
R is empty.
The set Q consists of the following terms:
f(x0, f(a, f(f(a, a), a)))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.