YES Termination w.r.t. Q proof of Strategy_removed_mixed_05_bn111.ari

(0) Obligation:

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

f(f(X, Y), Z) → f(X, f(Y, Z))
f(X, f(Y, Z)) → f(Y, 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(f(X, Y), Z) → F(X, f(Y, Z))
F(f(X, Y), Z) → F(Y, Z)
F(X, f(Y, Z)) → F(Y, Y)

The TRS R consists of the following rules:

f(f(X, Y), Z) → f(X, f(Y, Z))
f(X, f(Y, Z)) → f(Y, 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,JAR06].


The following pairs can be oriented strictly and are deleted.


F(f(X, Y), Z) → F(X, f(Y, Z))
F(f(X, Y), Z) → F(Y, Z)
F(X, f(Y, Z)) → F(Y, Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(F(x1, x2)) = x1 + 2·x1·x2 + 2·x12 + 2·x2   
POL(f(x1, x2)) = 2 + 2·x1 + x1·x2 + 2·x12 + 2·x2   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

f(X, f(Y, Z)) → f(Y, Y)
f(f(X, Y), Z) → f(X, f(Y, Z))

(4) Obligation:

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

f(f(X, Y), Z) → f(X, f(Y, Z))
f(X, f(Y, Z)) → f(Y, Y)

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

(5) PisEmptyProof (EQUIVALENT transformation)

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

(6) YES