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

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

(0) Obligation:

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

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

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

The TRS R consists of the following rules:

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

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

(3) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: Polynomial interpretation [POLO]:

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

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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: 0
f: x0
b: 1
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(f.0-1(a., b.), x) → F.0-0(a., f.0-0(a., x))
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
F.0-1(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)
F.0-1(x, f.1-1(y, z)) → F.0-1(f.0-1(x, y), z)
F.1-0(x, f.0-0(y, z)) → F.1-0(f.1-0(x, y), z)
F.1-0(x, f.0-1(y, z)) → F.1-1(f.1-0(x, y), z)
F.1-1(x, f.1-0(y, z)) → F.1-0(f.1-1(x, y), z)
F.1-1(x, f.1-1(y, z)) → F.1-1(f.1-1(x, y), z)
F.0-1(f.0-1(a., b.), x) → F.0-0(a., f.0-1(a., x))
F.1-0(f.1-0(b., a.), x) → F.1-1(b., f.1-0(b., x))
F.1-1(f.1-0(b., a.), x) → F.1-1(b., f.1-1(b., x))

The TRS R consists of the following rules:

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

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 2 SCCs.

(8) Complex Obligation (AND)

(9) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(10) 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.

The following dependency pairs can be deleted:

F.1-0(f.1-0(b., a.), x) → F.1-1(b., f.1-0(b., x))
F.1-0(x, f.0-0(y, z)) → F.1-0(f.1-0(x, y), z)
F.1-0(x, f.0-1(y, z)) → F.1-1(f.1-0(x, y), z)
F.1-1(f.1-0(b., a.), x) → F.1-1(b., f.1-1(b., x))
The following rules are removed from R:

f.1-0(f.1-0(b., a.), x) → f.1-1(b., f.1-0(b., x))
f.1-0(x, f.0-0(y, z)) → f.1-0(f.1-0(x, y), z)
f.1-1(f.1-0(b., a.), x) → f.1-1(b., f.1-1(b., x))
f.1-0(x, f.0-1(y, z)) → f.1-1(f.1-0(x, y), z)
Used ordering: POLO with Polynomial interpretation [POLO]:

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

(11) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(12) DependencyGraphProof (EQUIVALENT transformation)

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

(13) Obligation:

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

F.1-1(x, f.1-1(y, z)) → F.1-1(f.1-1(x, y), z)

The TRS R consists of the following rules:

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

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

(14) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F.1-1(x, f.1-1(y, z)) → F.1-1(f.1-1(x, y), z)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented: none

(15) Obligation:

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

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

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

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) TRUE

(18) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(19) 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.

The following dependency pairs can be deleted:

F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
F.0-1(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)
F.0-0(f.0-1(a., b.), x) → F.0-0(a., f.0-0(a., x))
F.0-1(x, f.1-1(y, z)) → F.0-1(f.0-1(x, y), z)
F.0-1(f.0-1(a., b.), x) → F.0-0(a., f.0-1(a., x))
The following rules are removed from R:

f.0-1(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.0-1(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.0-1(f.0-1(a., b.), x) → f.0-0(a., f.0-1(a., x))
f.0-0(f.0-1(a., b.), x) → f.0-0(a., f.0-0(a., x))
Used ordering: POLO with Polynomial interpretation [POLO]:

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

(20) Obligation:

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

F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)

The TRS R consists of the following rules:

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

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented: none

(22) Obligation:

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

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

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

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) TRUE