YES Termination w.r.t. Q proof of Secret_05_TRS_teparla2.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 3 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 SemLabProof (⇒, 69 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 QDP
↳9 UsableRulesReductionPairsProof (⇔, 0 ms)
↳10 QDP
↳11 DependencyGraphProof (⇔, 0 ms)
↳12 QDP
↳13 MRRProof (⇔, 0 ms)
↳14 QDP
↳15 UsableRulesReductionPairsProof (⇔, 0 ms)
↳16 QDP
↳17 QDPOrderProof (⇔, 0 ms)
↳18 QDP
↳19 PisEmptyProof (⇔, 0 ms)
↳20 YES

(0) Obligation:

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

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

The TRS R consists of the following rules:

f(y, f(x, f(a, x))) → f(f(f(a, x), 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 TRSs R and P. 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 QDP problem.

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

(13) 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 rules of the TRS R:

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

Used ordering: 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)) = 1 + x1 + x2   
POL(f.1-1(x1, x2)) = x1 + x2   

(14) 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.0-0(f.1-1(a., x), f.1-1(x, a.)), f.1-0(a., y))

The TRS R consists of the following rules:

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.

(15) 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.0-0(x, f.0-1(x, y)) → f.0-1(f.0-1(f.0-1(x, a.), a.), a.)
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.1-0(x1, x2)) = x1 + x2   
POL(f.1-1(x1, x2)) = x1 + x2   

(16) 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.0-0(f.1-1(a., x), f.1-1(x, a.)), f.1-0(a., y))

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.0-0(f.1-1(a., x), f.1-1(x, a.)), f.1-0(a., y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(18) Obligation:

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) YES