YES Termination w.r.t. Q proof of AProVE_04_Liveness6.3.ari

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 47 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 0 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 QDPSizeChangeProof (⇔, 0 ms)
        ↳28 YES
    ↳29 QDP
        ↳30 UsableRulesProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 QDPOrderProof (⇔, 10 ms)
        ↳33 QDP
        ↳34 PisEmptyProof (⇔, 0 ms)
        ↳35 YES

(0) Obligation:

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

rec(rec(x)) → sent(rec(x))
rec(sent(x)) → sent(rec(x))
rec(no(x)) → sent(rec(x))
rec(bot) → up(sent(bot))
rec(up(x)) → up(rec(x))
sent(up(x)) → up(sent(x))
no(up(x)) → up(no(x))
top(rec(up(x))) → top(check(rec(x)))
top(sent(up(x))) → top(check(rec(x)))
top(no(up(x))) → top(check(rec(x)))
check(up(x)) → up(check(x))
check(sent(x)) → sent(check(x))
check(rec(x)) → rec(check(x))
check(no(x)) → no(check(x))
check(no(x)) → no(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:

rec(rec(x)) → rec(sent(x))
sent(rec(x)) → rec(sent(x))
no(rec(x)) → rec(sent(x))
bot'(rec(x)) → bot'(sent(up(x)))
up(rec(x)) → rec(up(x))
up(sent(x)) → sent(up(x))
up(no(x)) → no(up(x))
up(rec(top(x))) → rec(check(top(x)))
up(sent(top(x))) → rec(check(top(x)))
up(no(top(x))) → rec(check(top(x)))
up(check(x)) → check(up(x))
sent(check(x)) → check(sent(x))
rec(check(x)) → check(rec(x))
no(check(x)) → check(no(x))
no(check(x)) → no(x)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(bot'(x1)) = x1   
POL(check(x1)) = x1   
POL(no(x1)) = 1 + x1   
POL(rec(x1)) = 1 + x1   
POL(sent(x1)) = x1   
POL(top(x1)) = x1   
POL(up(x1)) = 1 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

rec(rec(x)) → rec(sent(x))
no(rec(x)) → rec(sent(x))
up(rec(top(x))) → rec(check(top(x)))
up(no(top(x))) → rec(check(top(x)))


(4) Obligation:

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

sent(rec(x)) → rec(sent(x))
bot'(rec(x)) → bot'(sent(up(x)))
up(rec(x)) → rec(up(x))
up(sent(x)) → sent(up(x))
up(no(x)) → no(up(x))
up(sent(top(x))) → rec(check(top(x)))
up(check(x)) → check(up(x))
sent(check(x)) → check(sent(x))
rec(check(x)) → check(rec(x))
no(check(x)) → check(no(x))
no(check(x)) → no(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:

SENT(rec(x)) → REC(sent(x))
SENT(rec(x)) → SENT(x)
BOT'(rec(x)) → BOT'(sent(up(x)))
BOT'(rec(x)) → SENT(up(x))
BOT'(rec(x)) → UP(x)
UP(rec(x)) → REC(up(x))
UP(rec(x)) → UP(x)
UP(sent(x)) → SENT(up(x))
UP(sent(x)) → UP(x)
UP(no(x)) → NO(up(x))
UP(no(x)) → UP(x)
UP(sent(top(x))) → REC(check(top(x)))
UP(check(x)) → UP(x)
SENT(check(x)) → SENT(x)
REC(check(x)) → REC(x)
NO(check(x)) → NO(x)

The TRS R consists of the following rules:

sent(rec(x)) → rec(sent(x))
bot'(rec(x)) → bot'(sent(up(x)))
up(rec(x)) → rec(up(x))
up(sent(x)) → sent(up(x))
up(no(x)) → no(up(x))
up(sent(top(x))) → rec(check(top(x)))
up(check(x)) → check(up(x))
sent(check(x)) → check(sent(x))
rec(check(x)) → check(rec(x))
no(check(x)) → check(no(x))
no(check(x)) → no(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 5 SCCs with 7 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

NO(check(x)) → NO(x)

The TRS R consists of the following rules:

sent(rec(x)) → rec(sent(x))
bot'(rec(x)) → bot'(sent(up(x)))
up(rec(x)) → rec(up(x))
up(sent(x)) → sent(up(x))
up(no(x)) → no(up(x))
up(sent(top(x))) → rec(check(top(x)))
up(check(x)) → check(up(x))
sent(check(x)) → check(sent(x))
rec(check(x)) → check(rec(x))
no(check(x)) → check(no(x))
no(check(x)) → no(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:

NO(check(x)) → NO(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:

  • NO(check(x)) → NO(x)
    The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

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

REC(check(x)) → REC(x)

The TRS R consists of the following rules:

sent(rec(x)) → rec(sent(x))
bot'(rec(x)) → bot'(sent(up(x)))
up(rec(x)) → rec(up(x))
up(sent(x)) → sent(up(x))
up(no(x)) → no(up(x))
up(sent(top(x))) → rec(check(top(x)))
up(check(x)) → check(up(x))
sent(check(x)) → check(sent(x))
rec(check(x)) → check(rec(x))
no(check(x)) → check(no(x))
no(check(x)) → no(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:

REC(check(x)) → REC(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:

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

(18) YES

(19) Obligation:

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

SENT(check(x)) → SENT(x)
SENT(rec(x)) → SENT(x)

The TRS R consists of the following rules:

sent(rec(x)) → rec(sent(x))
bot'(rec(x)) → bot'(sent(up(x)))
up(rec(x)) → rec(up(x))
up(sent(x)) → sent(up(x))
up(no(x)) → no(up(x))
up(sent(top(x))) → rec(check(top(x)))
up(check(x)) → check(up(x))
sent(check(x)) → check(sent(x))
rec(check(x)) → check(rec(x))
no(check(x)) → check(no(x))
no(check(x)) → no(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:

SENT(check(x)) → SENT(x)
SENT(rec(x)) → SENT(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:

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

  • SENT(rec(x)) → SENT(x)
    The graph contains the following edges 1 > 1

(23) YES

(24) Obligation:

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

UP(sent(x)) → UP(x)
UP(rec(x)) → UP(x)
UP(no(x)) → UP(x)
UP(check(x)) → UP(x)

The TRS R consists of the following rules:

sent(rec(x)) → rec(sent(x))
bot'(rec(x)) → bot'(sent(up(x)))
up(rec(x)) → rec(up(x))
up(sent(x)) → sent(up(x))
up(no(x)) → no(up(x))
up(sent(top(x))) → rec(check(top(x)))
up(check(x)) → check(up(x))
sent(check(x)) → check(sent(x))
rec(check(x)) → check(rec(x))
no(check(x)) → check(no(x))
no(check(x)) → no(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:

UP(sent(x)) → UP(x)
UP(rec(x)) → UP(x)
UP(no(x)) → UP(x)
UP(check(x)) → UP(x)

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

(27) 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:

  • UP(sent(x)) → UP(x)
    The graph contains the following edges 1 > 1

  • UP(rec(x)) → UP(x)
    The graph contains the following edges 1 > 1

  • UP(no(x)) → UP(x)
    The graph contains the following edges 1 > 1

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

(28) YES

(29) Obligation:

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

BOT'(rec(x)) → BOT'(sent(up(x)))

The TRS R consists of the following rules:

sent(rec(x)) → rec(sent(x))
bot'(rec(x)) → bot'(sent(up(x)))
up(rec(x)) → rec(up(x))
up(sent(x)) → sent(up(x))
up(no(x)) → no(up(x))
up(sent(top(x))) → rec(check(top(x)))
up(check(x)) → check(up(x))
sent(check(x)) → check(sent(x))
rec(check(x)) → check(rec(x))
no(check(x)) → check(no(x))
no(check(x)) → no(x)

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

(30) 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.

(31) Obligation:

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

BOT'(rec(x)) → BOT'(sent(up(x)))

The TRS R consists of the following rules:

up(rec(x)) → rec(up(x))
up(sent(x)) → sent(up(x))
up(no(x)) → no(up(x))
up(sent(top(x))) → rec(check(top(x)))
up(check(x)) → check(up(x))
sent(rec(x)) → rec(sent(x))
sent(check(x)) → check(sent(x))
rec(check(x)) → check(rec(x))
no(check(x)) → check(no(x))
no(check(x)) → no(x)

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

(32) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


BOT'(rec(x)) → BOT'(sent(up(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(BOT'(x1)) = x1   
POL(check(x1)) = 0   
POL(no(x1)) = 1   
POL(rec(x1)) = 1 + x1   
POL(sent(x1)) = x1   
POL(top(x1)) = 1   
POL(up(x1)) = x1   

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

up(rec(x)) → rec(up(x))
up(sent(x)) → sent(up(x))
up(no(x)) → no(up(x))
up(sent(top(x))) → rec(check(top(x)))
up(check(x)) → check(up(x))
sent(rec(x)) → rec(sent(x))
sent(check(x)) → check(sent(x))
rec(check(x)) → check(rec(x))
no(check(x)) → check(no(x))
no(check(x)) → no(x)

(33) Obligation:

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

up(rec(x)) → rec(up(x))
up(sent(x)) → sent(up(x))
up(no(x)) → no(up(x))
up(sent(top(x))) → rec(check(top(x)))
up(check(x)) → check(up(x))
sent(rec(x)) → rec(sent(x))
sent(check(x)) → check(sent(x))
rec(check(x)) → check(rec(x))
no(check(x)) → check(no(x))
no(check(x)) → no(x)

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

(34) PisEmptyProof (EQUIVALENT transformation)

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

(35) YES