YES Termination w.r.t. Q proof of Secret_06_TRS_gen-15.ari

(0) Obligation:

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

c(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → z

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:

C(z, x, a) → F(b(b(f(z), z), x))
C(z, x, a) → B(b(f(z), z), x)
C(z, x, a) → B(f(z), z)
C(z, x, a) → F(z)
B(y, b(z, a)) → F(b(c(f(a), y, z), z))
B(y, b(z, a)) → B(c(f(a), y, z), z)
B(y, b(z, a)) → C(f(a), y, z)
B(y, b(z, a)) → F(a)

The TRS R consists of the following rules:

c(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → z

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 4 less nodes.

(4) Obligation:

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

C(z, x, a) → B(b(f(z), z), x)
B(y, b(z, a)) → B(c(f(a), y, z), z)
B(y, b(z, a)) → C(f(a), y, z)
C(z, x, a) → B(f(z), z)

The TRS R consists of the following rules:

c(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → z

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

(5) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule C(z, x, a) → B(b(f(z), z), x) we obtained the following new rules [LPAR04]:

C(f(a), y_0, a) → B(b(f(f(a)), f(a)), y_0) → C(f(a), y_0, a) → B(b(f(f(a)), f(a)), y_0)

(6) Obligation:

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

B(y, b(z, a)) → B(c(f(a), y, z), z)
B(y, b(z, a)) → C(f(a), y, z)
C(z, x, a) → B(f(z), z)
C(f(a), y_0, a) → B(b(f(f(a)), f(a)), y_0)

The TRS R consists of the following rules:

c(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → z

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

(7) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule C(z, x, a) → B(f(z), z) we obtained the following new rules [LPAR04]:

C(f(a), y_0, a) → B(f(f(a)), f(a)) → C(f(a), y_0, a) → B(f(f(a)), f(a))

(8) Obligation:

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

B(y, b(z, a)) → B(c(f(a), y, z), z)
B(y, b(z, a)) → C(f(a), y, z)
C(f(a), y_0, a) → B(b(f(f(a)), f(a)), y_0)
C(f(a), y_0, a) → B(f(f(a)), f(a))

The TRS R consists of the following rules:

c(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → z

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

(9) DependencyGraphProof (EQUIVALENT transformation)

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

(10) Obligation:

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

B(y, b(z, a)) → C(f(a), y, z)
C(f(a), y_0, a) → B(b(f(f(a)), f(a)), y_0)
B(y, b(z, a)) → B(c(f(a), y, z), z)

The TRS R consists of the following rules:

c(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → z

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

(11) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


C(f(a), y_0, a) → B(b(f(f(a)), f(a)), y_0)
B(y, b(z, a)) → B(c(f(a), y, z), z)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( B(x1, x2) ) = max{0, x1 + x2 - 1}

POL( c(x1, ..., x3) ) = 2x1 + x2 + 2

POL( a ) = 1

POL( f(x1) ) = max{0, x1 - 2}

POL( b(x1, x2) ) = x1 + x2 + 2

POL( C(x1, ..., x3) ) = x2 + x3 + 2


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

c(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → z

(12) Obligation:

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

B(y, b(z, a)) → C(f(a), y, z)

The TRS R consists of the following rules:

c(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → z

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

(13) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(14) TRUE