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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
  ↳5 SemLabProof (⇐)
    ↳6 QDP
    ↳7 DependencyGraphProof (⇔)
    ↳8 QDP
    ↳9 UsableRulesReductionPairsProof (⇔)
    ↳10 QDP
    ↳11 DependencyGraphProof (⇔)
    ↳12 QDP
  ↳13 SemLabProof2 (⇔)
    ↳14 QDP
    ↳15 RootLabelingFC2Proof (⇔)
    ↳16 QDP
    ↳17 DependencyGraphProof (⇔)
    ↳18 QDP
    ↳19 UsableRulesProof (⇔)
    ↳20 QDP
    ↳21 DependencyGraphProof (⇔)
    ↳22 QDP
    ↳23 QDPOrderProof (⇔)
    ↳24 QDP
    ↳25 PisEmptyProof (⇔)
    ↳26 TRUE

(0) Obligation:

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

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

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

The TRS R consists of the following rules:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 5 less nodes.

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(5) 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: 1
f: 0
F: 0
By semantic labelling [SEMLAB] we obtain the following labelled TRS.

(6) Obligation:

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

F.0-0(y, f.0-0(x, f.1-0(a., x))) → F.1-0(a., y)
F.0-0(y, f.0-0(x, f.1-0(a., x))) → F.0-0(f.1-0(a., f.0-1(x, a.)), f.1-0(a., y))
F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))
F.1-0(y, f.0-0(x, f.1-0(a., x))) → F.0-0(f.1-0(a., f.0-1(x, a.)), f.1-1(a., y))
F.1-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-1(a., y))
F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.1-0(a., y)
F.1-0(y, f.0-0(x, f.1-0(a., x))) → F.1-1(a., y)
F.1-0(y, f.1-0(x, f.1-1(a., x))) → F.1-1(a., y)

The TRS R consists of the following rules:

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

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 6 less nodes.

(8) Obligation:

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

F.0-0(y, f.0-0(x, f.1-0(a., x))) → F.0-0(f.1-0(a., f.0-1(x, a.)), f.1-0(a., y))
F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))

The TRS R consists of the following rules:

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

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.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(F.0-0(x1, x2)) = x1 + x2   
POL(a.) = 0   
POL(f.0-0(x1, x2)) = x1 + x2   
POL(f.0-1(x1, x2)) = x1 + x2   
POL(f.1-0(x1, x2)) = x1 + x2   
POL(f.1-1(x1, x2)) = x1 + x2   

(10) Obligation:

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

F.0-0(y, f.0-0(x, f.1-0(a., x))) → F.0-0(f.1-0(a., f.0-1(x, a.)), f.1-0(a., y))
F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))

The TRS R consists of the following rules:

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

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

(11) DependencyGraphProof (EQUIVALENT transformation)

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

(12) Obligation:

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

F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))

The TRS R consists of the following rules:

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

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

(13) SemLabProof2 (EQUIVALENT transformation)

As can be seen after transforming the QDP problem by semantic labelling [SEMLAB] and then some rule deleting processors, only certain labelled rules and pairs can be used. Hence, we only have to consider all unlabelled pairs and rules (without the decreasing rules for quasi-models).

(14) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(15) 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.

(16) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(18) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(19) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(20) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F_{f,f}(y, f_{a,f}(x, f_{a,a}(a, x))) → F_{f,f}(f_{a,f}(a, f_{a,a}(x, a)), f_{a,f}(a, y))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(F_{f,f}(x1, x2)) =
/0\
\1/
 +
/01\
\01/
·x1 +
/10\
\00/
·x2

POL(f_{a,f}(x1, x2)) =
/1\
\0/
 +
/01\
\00/
·x1 +
/01\
\00/
·x2

POL(f_{a,a}(x1, x2)) =
/0\
\1/
 +
/00\
\11/
·x1 +
/00\
\00/
·x2

POL(a) =
/1\
\1/

POL(f_{f,f}(x1, x2)) =
/0\
\0/
 +
/00\
\01/
·x1 +
/00\
\00/
·x2

POL(f_{f,a}(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/10\
\00/
·x2

The following usable rules [FROCOS05] were oriented:

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

(24) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

(25) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(26) TRUE