YES Termination w.r.t. Q proof of Transformed_CSR_04_Ex16_Luc06_iGM.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPSizeChangeProof (⇔, 0 ms)
        ↳9 YES
    ↳10 QDP
        ↳11 MRRProof (⇔, 0 ms)
        ↳12 QDP
        ↳13 TransformationProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 TransformationProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 DependencyGraphProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 UsableRulesProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 TransformationProof (⇔, 0 ms)
        ↳24 QDP
        ↳25 DependencyGraphProof (⇔, 0 ms)
        ↳26 TRUE

(0) Obligation:

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

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

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:

ACTIVE(f(X, X)) → MARK(f(a, b))
ACTIVE(f(X, X)) → F(a, b)
ACTIVE(b) → MARK(a)
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
MARK(f(X1, X2)) → F(mark(X1), X2)
MARK(f(X1, X2)) → MARK(X1)
MARK(a) → ACTIVE(a)
MARK(b) → ACTIVE(b)
F(mark(X1), X2) → F(X1, X2)
F(X1, mark(X2)) → F(X1, X2)
F(active(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

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 with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

F(X1, mark(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)
F(active(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

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

(6) 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.

(7) Obligation:

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

F(X1, mark(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)
F(active(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)

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

(8) 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:

  • F(X1, mark(X2)) → F(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

  • F(mark(X1), X2) → F(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • F(active(X1), X2) → F(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • F(X1, active(X2)) → F(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

(9) YES

(10) Obligation:

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

MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
ACTIVE(f(X, X)) → MARK(f(a, b))
MARK(f(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

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

(11) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

MARK(f(X1, X2)) → MARK(X1)


Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(a) = 0   
POL(active(x1)) = x1   
POL(b) = 0   
POL(f(x1, x2)) = 1 + 2·x1 + x2   
POL(mark(x1)) = x1   

(12) Obligation:

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

MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
ACTIVE(f(X, X)) → MARK(f(a, b))

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

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

(13) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2)) we obtained the following new rules [LPAR04]:

MARK(f(a, b)) → ACTIVE(f(mark(a), b)) → MARK(f(a, b)) → ACTIVE(f(mark(a), b))

(14) Obligation:

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

ACTIVE(f(X, X)) → MARK(f(a, b))
MARK(f(a, b)) → ACTIVE(f(mark(a), b))

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

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

(15) 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.

(16) Obligation:

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

ACTIVE(f(X, X)) → MARK(f(a, b))
MARK(f(a, b)) → ACTIVE(f(mark(a), b))

The TRS R consists of the following rules:

mark(a) → active(a)
f(X1, mark(X2)) → f(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

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

(17) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule MARK(f(a, b)) → ACTIVE(f(mark(a), b)) at position [0] we obtained the following new rules [LPAR04]:

MARK(f(a, b)) → ACTIVE(f(a, b)) → MARK(f(a, b)) → ACTIVE(f(a, b))
MARK(f(a, b)) → ACTIVE(f(active(a), b)) → MARK(f(a, b)) → ACTIVE(f(active(a), b))

(18) Obligation:

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

ACTIVE(f(X, X)) → MARK(f(a, b))
MARK(f(a, b)) → ACTIVE(f(a, b))
MARK(f(a, b)) → ACTIVE(f(active(a), b))

The TRS R consists of the following rules:

mark(a) → active(a)
f(X1, mark(X2)) → f(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

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

(19) DependencyGraphProof (EQUIVALENT transformation)

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

(20) Obligation:

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

MARK(f(a, b)) → ACTIVE(f(active(a), b))
ACTIVE(f(X, X)) → MARK(f(a, b))

The TRS R consists of the following rules:

mark(a) → active(a)
f(X1, mark(X2)) → f(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

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

(21) 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.

(22) Obligation:

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

MARK(f(a, b)) → ACTIVE(f(active(a), b))
ACTIVE(f(X, X)) → MARK(f(a, b))

The TRS R consists of the following rules:

f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

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

(23) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule MARK(f(a, b)) → ACTIVE(f(active(a), b)) at position [0] we obtained the following new rules [LPAR04]:

MARK(f(a, b)) → ACTIVE(f(a, b)) → MARK(f(a, b)) → ACTIVE(f(a, b))

(24) Obligation:

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

ACTIVE(f(X, X)) → MARK(f(a, b))
MARK(f(a, b)) → ACTIVE(f(a, b))

The TRS R consists of the following rules:

f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

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

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(26) TRUE