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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 PisEmptyProof (⇔)
        ↳14 TRUE

(0) Obligation:

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

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

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

The TRS R consists of the following rules:

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

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 2 SCCs.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

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

The following usable rules [FROCOS05] were oriented:

f(b, f(a, x)) → f(b, f(b, f(b, x)))

(7) Obligation:

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

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

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

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

The following usable rules [FROCOS05] were oriented:

f(a, f(b, x)) → f(a, f(a, f(a, x)))

(12) Obligation:

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

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

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE