YES Termination w.r.t. Q proof of Zantema_05_jw09.ari

(0) Obligation:

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

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

The TRS R consists of the following rules:

f(a, f(a, x)) → f(x, f(a, f(f(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 3 less nodes.

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(5) 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.

(6) Obligation:

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

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

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

(7) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


F(a, f(a, x)) → F(x, f(a, f(f(a, a), a)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( F(x1, x2) ) = x1 + x2

POL( a ) = 2

POL( f(x1, x2) ) = max{0, x2 - 1}


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

(8) Obligation:

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

(9) PisEmptyProof (EQUIVALENT transformation)

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

(10) YES