YES Termination w.r.t. Q proof of Der95_07.ari

(0) Obligation:

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

w(r(x)) → r(w(x))
b(r(x)) → r(b(x))
b(w(x)) → w(b(x))

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

r(w(x)) → w(r(x))
r(b(x)) → b(r(x))
w(b(x)) → b(w(x))

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

R(w(x)) → W(r(x))
R(w(x)) → R(x)
R(b(x)) → R(x)
W(b(x)) → W(x)

The TRS R consists of the following rules:

r(w(x)) → w(r(x))
r(b(x)) → b(r(x))
w(b(x)) → b(w(x))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

W(b(x)) → W(x)

The TRS R consists of the following rules:

r(w(x)) → w(r(x))
r(b(x)) → b(r(x))
w(b(x)) → b(w(x))

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

(8) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(9) Obligation:

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

W(b(x)) → W(x)

The TRS R consists of the following rules:

r(w(x)) → w(r(x))
r(b(x)) → b(r(x))
w(b(x)) → b(w(x))

The set Q consists of the following terms:

r(w(x0))
r(b(x0))
w(b(x0))

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

(10) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(11) Obligation:

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

W(b(x)) → W(x)

R is empty.
The set Q consists of the following terms:

r(w(x0))
r(b(x0))
w(b(x0))

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

(12) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

r(w(x0))
r(b(x0))
w(b(x0))

(13) Obligation:

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

W(b(x)) → W(x)

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

(14) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • W(b(x)) → W(x)
    The graph contains the following edges 1 > 1

(15) YES

(16) Obligation:

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

R(b(x)) → R(x)
R(w(x)) → R(x)

The TRS R consists of the following rules:

r(w(x)) → w(r(x))
r(b(x)) → b(r(x))
w(b(x)) → b(w(x))

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

(17) 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.

(18) Obligation:

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

R(b(x)) → R(x)
R(w(x)) → R(x)

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • R(b(x)) → R(x)
    The graph contains the following edges 1 > 1

  • R(w(x)) → R(x)
    The graph contains the following edges 1 > 1

(20) YES