YES Termination w.r.t. Q proof of AProVE_04_Liveness6.1.ari

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 38 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 8 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 UsableRulesProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QDPSizeChangeProof (⇔, 0 ms)
        ↳18 YES
    ↳19 QDP
        ↳20 UsableRulesProof (⇔, 0 ms)
        ↳21 QDP
        ↳22 QDPSizeChangeProof (⇔, 0 ms)
        ↳23 YES
    ↳24 QDP
        ↳25 UsableRulesProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 QDPOrderProof (⇔, 10 ms)
        ↳28 QDP
        ↳29 PisEmptyProof (⇔, 0 ms)
        ↳30 YES

(0) Obligation:

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

top(free(x)) → top(check(new(x)))
new(free(x)) → free(new(x))
old(free(x)) → free(old(x))
new(serve) → free(serve)
old(serve) → free(serve)
check(free(x)) → free(check(x))
check(new(x)) → new(check(x))
check(old(x)) → old(check(x))
check(old(x)) → old(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:

free(top(x)) → new(check(top(x)))
free(new(x)) → new(free(x))
free(old(x)) → old(free(x))
serve'(new(x)) → serve'(free(x))
serve'(old(x)) → serve'(free(x))
free(check(x)) → check(free(x))
new(check(x)) → check(new(x))
old(check(x)) → check(old(x))
old(check(x)) → old(x)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(check(x1)) = x1   
POL(free(x1)) = x1   
POL(new(x1)) = x1   
POL(old(x1)) = 1 + x1   
POL(serve'(x1)) = x1   
POL(top(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

serve'(old(x)) → serve'(free(x))


(4) Obligation:

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

free(top(x)) → new(check(top(x)))
free(new(x)) → new(free(x))
free(old(x)) → old(free(x))
serve'(new(x)) → serve'(free(x))
free(check(x)) → check(free(x))
new(check(x)) → check(new(x))
old(check(x)) → check(old(x))
old(check(x)) → old(x)

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

FREE(top(x)) → NEW(check(top(x)))
FREE(new(x)) → NEW(free(x))
FREE(new(x)) → FREE(x)
FREE(old(x)) → OLD(free(x))
FREE(old(x)) → FREE(x)
SERVE'(new(x)) → SERVE'(free(x))
SERVE'(new(x)) → FREE(x)
FREE(check(x)) → FREE(x)
NEW(check(x)) → NEW(x)
OLD(check(x)) → OLD(x)

The TRS R consists of the following rules:

free(top(x)) → new(check(top(x)))
free(new(x)) → new(free(x))
free(old(x)) → old(free(x))
serve'(new(x)) → serve'(free(x))
free(check(x)) → check(free(x))
new(check(x)) → check(new(x))
old(check(x)) → check(old(x))
old(check(x)) → old(x)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

OLD(check(x)) → OLD(x)

The TRS R consists of the following rules:

free(top(x)) → new(check(top(x)))
free(new(x)) → new(free(x))
free(old(x)) → old(free(x))
serve'(new(x)) → serve'(free(x))
free(check(x)) → check(free(x))
new(check(x)) → check(new(x))
old(check(x)) → check(old(x))
old(check(x)) → old(x)

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

(10) 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.

(11) Obligation:

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

OLD(check(x)) → OLD(x)

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

(12) 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:

  • OLD(check(x)) → OLD(x)
    The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

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

NEW(check(x)) → NEW(x)

The TRS R consists of the following rules:

free(top(x)) → new(check(top(x)))
free(new(x)) → new(free(x))
free(old(x)) → old(free(x))
serve'(new(x)) → serve'(free(x))
free(check(x)) → check(free(x))
new(check(x)) → check(new(x))
old(check(x)) → check(old(x))
old(check(x)) → old(x)

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:

NEW(check(x)) → NEW(x)

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

(17) 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:

  • NEW(check(x)) → NEW(x)
    The graph contains the following edges 1 > 1

(18) YES

(19) Obligation:

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

FREE(old(x)) → FREE(x)
FREE(new(x)) → FREE(x)
FREE(check(x)) → FREE(x)

The TRS R consists of the following rules:

free(top(x)) → new(check(top(x)))
free(new(x)) → new(free(x))
free(old(x)) → old(free(x))
serve'(new(x)) → serve'(free(x))
free(check(x)) → check(free(x))
new(check(x)) → check(new(x))
old(check(x)) → check(old(x))
old(check(x)) → old(x)

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

(20) 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.

(21) Obligation:

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

FREE(old(x)) → FREE(x)
FREE(new(x)) → FREE(x)
FREE(check(x)) → FREE(x)

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

(22) 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:

  • FREE(old(x)) → FREE(x)
    The graph contains the following edges 1 > 1

  • FREE(new(x)) → FREE(x)
    The graph contains the following edges 1 > 1

  • FREE(check(x)) → FREE(x)
    The graph contains the following edges 1 > 1

(23) YES

(24) Obligation:

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

SERVE'(new(x)) → SERVE'(free(x))

The TRS R consists of the following rules:

free(top(x)) → new(check(top(x)))
free(new(x)) → new(free(x))
free(old(x)) → old(free(x))
serve'(new(x)) → serve'(free(x))
free(check(x)) → check(free(x))
new(check(x)) → check(new(x))
old(check(x)) → check(old(x))
old(check(x)) → old(x)

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

(25) 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.

(26) Obligation:

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

SERVE'(new(x)) → SERVE'(free(x))

The TRS R consists of the following rules:

free(top(x)) → new(check(top(x)))
free(new(x)) → new(free(x))
free(old(x)) → old(free(x))
free(check(x)) → check(free(x))
old(check(x)) → check(old(x))
old(check(x)) → old(x)
new(check(x)) → check(new(x))

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

(27) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


SERVE'(new(x)) → SERVE'(free(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(SERVE'(x1)) = x1   
POL(check(x1)) = 0   
POL(free(x1)) = x1   
POL(new(x1)) = 1 + x1   
POL(old(x1)) = 0   
POL(top(x1)) = 1   

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

free(top(x)) → new(check(top(x)))
free(new(x)) → new(free(x))
free(old(x)) → old(free(x))
free(check(x)) → check(free(x))
old(check(x)) → check(old(x))
old(check(x)) → old(x)
new(check(x)) → check(new(x))

(28) Obligation:

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

free(top(x)) → new(check(top(x)))
free(new(x)) → new(free(x))
free(old(x)) → old(free(x))
free(check(x)) → check(free(x))
old(check(x)) → check(old(x))
old(check(x)) → old(x)
new(check(x)) → check(new(x))

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

(29) PisEmptyProof (EQUIVALENT transformation)

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

(30) YES