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, a(b(y))) → f(a(b(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
Q is empty.
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, a(b(y))) → f(a(b(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
Q is empty.
We have applied [19,8] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
f(x, a(b(y))) → f(a(b(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
The signature Sigma is {f}
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, a(b(y))) → f(a(b(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
The set Q consists of the following terms:
f(x0, a(b(x1)))
f(a(x0), x1)
f(b(x0), x1)
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(a(x), y) → F(x, a(y))
F(x, a(b(y))) → F(a(b(x)), y)
F(b(x), y) → F(x, b(y))
The TRS R consists of the following rules:
f(x, a(b(y))) → f(a(b(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
The set Q consists of the following terms:
f(x0, a(b(x1)))
f(a(x0), x1)
f(b(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(a(x), y) → F(x, a(y))
F(x, a(b(y))) → F(a(b(x)), y)
F(b(x), y) → F(x, b(y))
The TRS R consists of the following rules:
f(x, a(b(y))) → f(a(b(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
The set Q consists of the following terms:
f(x0, a(b(x1)))
f(a(x0), x1)
f(b(x0), x1)
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
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
F(a(x), y) → F(x, a(y))
F(x, a(b(y))) → F(a(b(x)), y)
F(b(x), y) → F(x, b(y))
R is empty.
The set Q consists of the following terms:
f(x0, a(b(x1)))
f(a(x0), x1)
f(b(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
f(x0, a(b(x1)))
f(a(x0), x1)
f(b(x0), x1)
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPToSRSProof
Q DP problem:
The TRS P consists of the following rules:
F(a(x), y) → F(x, a(y))
F(x, a(b(y))) → F(a(b(x)), y)
F(b(x), y) → F(x, b(y))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(a(x))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(x)
The TRS R consists of the following rules:
a(b(x)) → b(a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(x)
The TRS R consists of the following rules:
a(b(x)) → b(a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(x)
The TRS R consists of the following rules:
a(b(x)) → b(a(x))
The set Q consists of the following terms:
a(b(x0))
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
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(x)
R is empty.
The set Q consists of the following terms:
a(b(x0))
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
a(b(x0))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs:
- A(b(x)) → A(x)
The graph contains the following edges 1 > 1