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(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(b, x)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(b, 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:
F(b, f(a, x)) → F(b, f(b, f(b, x)))
F(a, f(b, x)) → F(a, f(a, f(a, x)))
F(a, f(b, x)) → F(a, f(a, x))
F(b, f(a, x)) → F(b, x)
F(a, f(b, x)) → F(a, x)
F(b, f(a, x)) → F(b, f(b, x))
The TRS R consists of the following rules:
f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(b, 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:
F(b, f(a, x)) → F(b, f(b, f(b, x)))
F(a, f(b, x)) → F(a, f(a, f(a, x)))
F(a, f(b, x)) → F(a, f(a, x))
F(b, f(a, x)) → F(b, x)
F(a, f(b, x)) → F(a, x)
F(b, f(a, x)) → F(b, f(b, x))
The TRS R consists of the following rules:
f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(b, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(b, f(a, x)) → F(b, f(b, f(b, x)))
F(b, f(a, x)) → F(b, x)
F(b, f(a, x)) → F(b, f(b, x))
The TRS R consists of the following rules:
f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(b, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(b, f(a, x)) → F(b, f(b, f(b, x)))
F(b, f(a, x)) → F(b, x)
F(b, f(a, x)) → F(b, f(b, x))
The TRS R consists of the following rules:
f(b, f(a, x)) → f(b, f(b, f(b, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(b, f(a, x)) → F(b, x)
The TRS R consists of the following rules:
f(b, f(a, x)) → f(b, f(b, f(b, 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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(b, f(a, x)) → F(b, x)
The TRS R consists of the following rules:
f(b, f(a, x)) → f(b, f(b, f(b, x)))
The set Q consists of the following terms:
f(b, f(a, 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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(b, f(a, x)) → F(b, x)
R is empty.
The set Q consists of the following terms:
f(b, f(a, x0))
We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
b1(a(x)) → b1(x)
R is empty.
The set Q consists of the following terms:
b(a(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.
b(a(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
b1(a(x)) → b1(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:
- b1(a(x)) → b1(x)
The graph contains the following edges 1 > 1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(a, f(b, x)) → F(a, f(a, f(a, x)))
F(a, f(b, x)) → F(a, f(a, x))
F(a, f(b, x)) → F(a, x)
The TRS R consists of the following rules:
f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(b, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(a, f(b, x)) → F(a, f(a, f(a, x)))
F(a, f(b, x)) → F(a, f(a, x))
F(a, f(b, x)) → F(a, x)
The TRS R consists of the following rules:
f(a, f(b, x)) → f(a, f(a, f(a, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
F(a, f(b, x)) → F(a, x)
The TRS R consists of the following rules:
f(a, f(b, x)) → f(a, f(a, f(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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(a, f(b, x)) → F(a, x)
The TRS R consists of the following rules:
f(a, f(b, x)) → f(a, f(a, f(a, x)))
The set Q consists of the following terms:
f(a, f(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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
Q DP problem:
The TRS P consists of the following rules:
F(a, f(b, x)) → F(a, x)
R is empty.
The set Q consists of the following terms:
f(a, f(b, x0))
We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
a1(b(x)) → a1(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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
Q DP problem:
The TRS P consists of the following rules:
a1(b(x)) → a1(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:
- a1(b(x)) → a1(x)
The graph contains the following edges 1 > 1