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, f(a, x))) → f(a, f(b, f(b, f(a, x))))
f(b, f(b, f(b, x))) → 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, f(a, x))) → f(a, f(b, f(b, f(a, x))))
f(b, f(b, f(b, x))) → 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(a, f(b, f(a, x))) → F(b, f(b, f(a, x)))
F(a, f(b, f(a, x))) → F(a, f(b, f(b, f(a, x))))
The TRS R consists of the following rules:
f(a, f(b, f(a, x))) → f(a, f(b, f(b, f(a, x))))
f(b, f(b, f(b, x))) → 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(a, f(b, f(a, x))) → F(b, f(b, f(a, x)))
F(a, f(b, f(a, x))) → F(a, f(b, f(b, f(a, x))))
The TRS R consists of the following rules:
f(a, f(b, f(a, x))) → f(a, f(b, f(b, f(a, x))))
f(b, f(b, f(b, x))) → 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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesReductionPairsProof
Q DP problem:
The TRS P consists of the following rules:
F(a, f(b, f(a, x))) → F(a, f(b, f(b, f(a, x))))
The TRS R consists of the following rules:
f(a, f(b, f(a, x))) → f(a, f(b, f(b, f(a, x))))
f(b, f(b, f(b, x))) → f(b, f(b, x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
First, we A-transformed [17] the QDP-Problem.
Then we obtain the following A-transformed DP problem.
The pairs P are:
a1(b(a(x))) → a1(b(b(a(x))))
and the Q and R are:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(a(x))) → a(b(b(a(x))))
b(b(b(x))) → b(b(x))
Q is empty.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
No dependency pairs are removed.
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(a(x1)) = x1
POL(a1(x1)) = x1
POL(b(x1)) = x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ RFCMatchBoundsDPProof
Q DP problem:
The TRS P consists of the following rules:
a1(b(a(x))) → a1(b(b(a(x))))
The TRS R consists of the following rules:
a(b(a(x))) → a(b(b(a(x))))
b(b(b(x))) → b(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
Finiteness of the DP problem can be shown by a matchbound of 1.
As the DP problem is minimal we only have to initialize the certificate graph by the rules of P:
a1(b(a(x))) → a1(b(b(a(x))))
To find matches we regarded all rules of R and P:
a(b(a(x))) → a(b(b(a(x))))
b(b(b(x))) → b(b(x))
a1(b(a(x))) → a1(b(b(a(x))))
The certificate found is represented by the following graph.
The certificate consists of the following enumerated nodes:
28, 29, 30, 31, 32, 34, 35, 33
Node 28 is start node and node 29 is final node.
Those nodes are connect through the following edges:
- 28 to 30 labelled a1_1(0)
- 29 to 29 labelled #_1(0)
- 30 to 31 labelled b_1(0)
- 31 to 32 labelled b_1(0)
- 32 to 29 labelled a_1(0)
- 32 to 33 labelled a_1(1)
- 34 to 35 labelled b_1(1)
- 35 to 29 labelled a_1(1)
- 35 to 33 labelled a_1(1)
- 33 to 34 labelled b_1(1)