Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 QDPOrderProof (⇔)
↳4 QDP
↳5 RootLabelingFC2Proof (⇔)
↳6 QDP
↳7 DependencyGraphProof (⇔)
↳8 QDP
↳9 UsableRulesReductionPairsProof (⇔)
↳10 QDP
↳11 SemLabProof (⇐)
↳12 QDP
↳13 DependencyGraphProof (⇔)
↳14 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(x, f(y, x)) → f(f(x, x), f(a, y))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(x, f(y, x)) → F(f(x, x), f(a, y))
F(x, f(y, x)) → F(x, x)
F(x, f(y, x)) → F(a, y)

The TRS R consists of the following rules:

f(x, f(y, x)) → f(f(x, x), f(a, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(x, f(y, x)) → F(x, x)
F(x, f(y, x)) → F(a, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(F(x1, x2)) = 0A + 1A·x1 + 1A·x2

POL(f(x1, x2)) = 5A + 1A·x1 + 1A·x2

POL(a) = 4A

The following usable rules [FROCOS05] were oriented:

f(x, f(y, x)) → f(f(x, x), f(a, y))

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(x, f(y, x)) → F(f(x, x), f(a, y))

The TRS R consists of the following rules:

f(x, f(y, x)) → f(f(x, x), f(a, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) RootLabelingFC2Proof (EQUIVALENT transformation)

We used root labeling (second transformation) [ROOTLAB] with the following heuristic:
LabelAll: All function symbols get labeled
As Q is empty the root labeling was sound AND complete.

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F_{f,f}(x, f_{f,f}(y, x)) → F_{f,f}(f_{f,f}(x, x), f_{a,f}(a, y))
F_{f,f}(x, f_{a,f}(y, x)) → F_{f,f}(f_{f,f}(x, x), f_{a,a}(a, y))
F_{a,f}(x, f_{f,a}(y, x)) → F_{f,f}(f_{a,a}(x, x), f_{a,f}(a, y))
F_{a,f}(x, f_{a,a}(y, x)) → F_{f,f}(f_{a,a}(x, x), f_{a,a}(a, y))

The TRS R consists of the following rules:

f_{f,f}(x, f_{f,f}(y, x)) → f_{f,f}(f_{f,f}(x, x), f_{a,f}(a, y))
f_{f,f}(x, f_{a,f}(y, x)) → f_{f,f}(f_{f,f}(x, x), f_{a,a}(a, y))
f_{a,f}(x, f_{f,a}(y, x)) → f_{f,f}(f_{a,a}(x, x), f_{a,f}(a, y))
f_{a,f}(x, f_{a,a}(y, x)) → f_{f,f}(f_{a,a}(x, x), f_{a,a}(a, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F_{f,f}(x, f_{f,f}(y, x)) → F_{f,f}(f_{f,f}(x, x), f_{a,f}(a, y))

The TRS R consists of the following rules:

f_{f,f}(x, f_{f,f}(y, x)) → f_{f,f}(f_{f,f}(x, x), f_{a,f}(a, y))
f_{f,f}(x, f_{a,f}(y, x)) → f_{f,f}(f_{f,f}(x, x), f_{a,a}(a, y))
f_{a,f}(x, f_{f,a}(y, x)) → f_{f,f}(f_{a,a}(x, x), f_{a,f}(a, y))
f_{a,f}(x, f_{a,a}(y, x)) → f_{f,f}(f_{a,a}(x, x), f_{a,a}(a, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(9) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] 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.

The following rules are removed from R:

f_{a,f}(x, f_{f,a}(y, x)) → f_{f,f}(f_{a,a}(x, x), f_{a,f}(a, y))
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(F_{f,f}(x1, x2)) = 2·x1 + 2·x2   
POL(a) = 0   
POL(f_{a,a}(x1, x2)) = 2·x1 + x2   
POL(f_{a,f}(x1, x2)) = 2·x1 + x2   
POL(f_{f,a}(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(f_{f,f}(x1, x2)) = x1 + 2·x2   

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F_{f,f}(x, f_{f,f}(y, x)) → F_{f,f}(f_{f,f}(x, x), f_{a,f}(a, y))

The TRS R consists of the following rules:

f_{f,f}(x, f_{f,f}(y, x)) → f_{f,f}(f_{f,f}(x, x), f_{a,f}(a, y))
f_{f,f}(x, f_{a,f}(y, x)) → f_{f,f}(f_{f,f}(x, x), f_{a,a}(a, y))
f_{a,f}(x, f_{a,a}(y, x)) → f_{f,f}(f_{a,a}(x, x), f_{a,a}(a, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(11) SemLabProof (SOUND transformation)

We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.a: 0
f_{f,f}: 0
f_{a,a}: 1
f_{a,f}: 0
F_{f,f}: 0
By semantic labelling [SEMLAB] we obtain the following labelled TRS.

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F_{f,f}.0-0(x, f_{f,f}.0-0(y, x)) → F_{f,f}.0-0(f_{f,f}.0-0(x, x), f_{a,f}.0-0(a., y))
F_{f,f}.0-0(x, f_{f,f}.1-0(y, x)) → F_{f,f}.0-0(f_{f,f}.0-0(x, x), f_{a,f}.0-1(a., y))
F_{f,f}.1-0(x, f_{f,f}.0-1(y, x)) → F_{f,f}.0-0(f_{f,f}.1-1(x, x), f_{a,f}.0-0(a., y))
F_{f,f}.1-0(x, f_{f,f}.1-1(y, x)) → F_{f,f}.0-0(f_{f,f}.1-1(x, x), f_{a,f}.0-1(a., y))

The TRS R consists of the following rules:

f_{f,f}.1-0(x, f_{f,f}.0-1(y, x)) → f_{f,f}.0-0(f_{f,f}.1-1(x, x), f_{a,f}.0-0(a., y))
f_{f,f}.1-0(x, f_{a,f}.1-1(y, x)) → f_{f,f}.0-1(f_{f,f}.1-1(x, x), f_{a,a}.0-1(a., y))
f_{a,f}.0-1(x, f_{a,a}.0-0(y, x)) → f_{f,f}.1-1(f_{a,a}.0-0(x, x), f_{a,a}.0-0(a., y))
f_{f,f}.1-0(x, f_{a,f}.0-1(y, x)) → f_{f,f}.0-1(f_{f,f}.1-1(x, x), f_{a,a}.0-0(a., y))
f_{f,f}.0-0(x, f_{a,f}.1-0(y, x)) → f_{f,f}.0-1(f_{f,f}.0-0(x, x), f_{a,a}.0-1(a., y))
f_{f,f}.0-0(x, f_{a,f}.0-0(y, x)) → f_{f,f}.0-1(f_{f,f}.0-0(x, x), f_{a,a}.0-0(a., y))
f_{f,f}.1-0(x, f_{f,f}.1-1(y, x)) → f_{f,f}.0-0(f_{f,f}.1-1(x, x), f_{a,f}.0-1(a., y))
f_{a,f}.1-1(x, f_{a,a}.1-1(y, x)) → f_{f,f}.1-1(f_{a,a}.1-1(x, x), f_{a,a}.0-1(a., y))
f_{f,f}.0-0(x, f_{f,f}.0-0(y, x)) → f_{f,f}.0-0(f_{f,f}.0-0(x, x), f_{a,f}.0-0(a., y))
f_{a,f}.1-1(x, f_{a,a}.0-1(y, x)) → f_{f,f}.1-1(f_{a,a}.1-1(x, x), f_{a,a}.0-0(a., y))
f_{a,f}.0-1(x, f_{a,a}.1-0(y, x)) → f_{f,f}.1-1(f_{a,a}.0-0(x, x), f_{a,a}.0-1(a., y))
f_{f,f}.0-0(x, f_{f,f}.1-0(y, x)) → f_{f,f}.0-0(f_{f,f}.0-0(x, x), f_{a,f}.0-1(a., y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 4 less nodes.

(14) TRUE