YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z085-rotate.srs

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

0 QTRS
↳1 QTRSRRRProof (⇔, 66 ms)
↳2 QTRS
↳3 Overlay + Local Confluence (⇔, 21 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 0 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 UsableRulesProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QReductionProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 QDPSizeChangeProof (⇔, 0 ms)
        ↳15 YES
    ↳16 QDP
        ↳17 UsableRulesProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QReductionProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 QDPOrderProof (⇔, 20 ms)
        ↳22 QDP
        ↳23 QDPOrderProof (⇔, 17 ms)
        ↳24 QDP
        ↳25 PisEmptyProof (⇔, 0 ms)
        ↳26 YES
    ↳27 QDP
        ↳28 UsableRulesProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 QReductionProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 QDPSizeChangeProof (⇔, 0 ms)
        ↳33 YES

(0) Obligation:

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

begin(end(x)) → rewrite(end(x))
begin(a(x)) → rotate(cut(Ca(guess(x))))
begin(b(x)) → rotate(cut(Cb(guess(x))))
guess(a(x)) → Ca(guess(x))
guess(b(x)) → Cb(guess(x))
guess(a(x)) → moveleft(Ba(wait(x)))
guess(b(x)) → moveleft(Bb(wait(x)))
guess(end(x)) → finish(end(x))
Ca(moveleft(Ba(x))) → moveleft(Ba(Aa(x)))
Cb(moveleft(Ba(x))) → moveleft(Ba(Ab(x)))
Ca(moveleft(Bb(x))) → moveleft(Bb(Aa(x)))
Cb(moveleft(Bb(x))) → moveleft(Bb(Ab(x)))
cut(moveleft(Ba(x))) → Da(cut(goright(x)))
cut(moveleft(Bb(x))) → Db(cut(goright(x)))
goright(Aa(x)) → Ca(goright(x))
goright(Ab(x)) → Cb(goright(x))
goright(wait(a(x))) → moveleft(Ba(wait(x)))
goright(wait(b(x))) → moveleft(Bb(wait(x)))
goright(wait(end(x))) → finish(end(x))
Ca(finish(x)) → finish(a(x))
Cb(finish(x)) → finish(b(x))
cut(finish(x)) → finish2(x)
Da(finish2(x)) → finish2(a(x))
Db(finish2(x)) → finish2(b(x))
rotate(finish2(x)) → rewrite(x)
rewrite(a(x)) → begin(b(b(x)))
rewrite(b(b(b(x)))) → begin(a(x))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = 30 + x1   
POL(Ab(x1)) = 12 + x1   
POL(Ba(x1)) = 30 + x1   
POL(Bb(x1)) = 12 + x1   
POL(Ca(x1)) = 30 + x1   
POL(Cb(x1)) = 12 + x1   
POL(Da(x1)) = 30 + x1   
POL(Db(x1)) = 12 + x1   
POL(a(x1)) = 30 + x1   
POL(b(x1)) = 12 + x1   
POL(begin(x1)) = 5 + x1   
POL(cut(x1)) = 2 + x1   
POL(end(x1)) = x1   
POL(finish(x1)) = x1   
POL(finish2(x1)) = 1 + x1   
POL(goright(x1)) = x1   
POL(guess(x1)) = 2 + x1   
POL(moveleft(x1)) = x1   
POL(rewrite(x1)) = x1   
POL(rotate(x1)) = x1   
POL(wait(x1)) = 1 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

begin(end(x)) → rewrite(end(x))
begin(a(x)) → rotate(cut(Ca(guess(x))))
begin(b(x)) → rotate(cut(Cb(guess(x))))
guess(a(x)) → moveleft(Ba(wait(x)))
guess(b(x)) → moveleft(Bb(wait(x)))
guess(end(x)) → finish(end(x))
goright(wait(end(x))) → finish(end(x))
cut(finish(x)) → finish2(x)
rotate(finish2(x)) → rewrite(x)
rewrite(a(x)) → begin(b(b(x)))
rewrite(b(b(b(x)))) → begin(a(x))


(2) Obligation:

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

guess(a(x)) → Ca(guess(x))
guess(b(x)) → Cb(guess(x))
Ca(moveleft(Ba(x))) → moveleft(Ba(Aa(x)))
Cb(moveleft(Ba(x))) → moveleft(Ba(Ab(x)))
Ca(moveleft(Bb(x))) → moveleft(Bb(Aa(x)))
Cb(moveleft(Bb(x))) → moveleft(Bb(Ab(x)))
cut(moveleft(Ba(x))) → Da(cut(goright(x)))
cut(moveleft(Bb(x))) → Db(cut(goright(x)))
goright(Aa(x)) → Ca(goright(x))
goright(Ab(x)) → Cb(goright(x))
goright(wait(a(x))) → moveleft(Ba(wait(x)))
goright(wait(b(x))) → moveleft(Bb(wait(x)))
Ca(finish(x)) → finish(a(x))
Cb(finish(x)) → finish(b(x))
Da(finish2(x)) → finish2(a(x))
Db(finish2(x)) → finish2(b(x))

Q is empty.

(3) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(4) Obligation:

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

guess(a(x)) → Ca(guess(x))
guess(b(x)) → Cb(guess(x))
Ca(moveleft(Ba(x))) → moveleft(Ba(Aa(x)))
Cb(moveleft(Ba(x))) → moveleft(Ba(Ab(x)))
Ca(moveleft(Bb(x))) → moveleft(Bb(Aa(x)))
Cb(moveleft(Bb(x))) → moveleft(Bb(Ab(x)))
cut(moveleft(Ba(x))) → Da(cut(goright(x)))
cut(moveleft(Bb(x))) → Db(cut(goright(x)))
goright(Aa(x)) → Ca(goright(x))
goright(Ab(x)) → Cb(goright(x))
goright(wait(a(x))) → moveleft(Ba(wait(x)))
goright(wait(b(x))) → moveleft(Bb(wait(x)))
Ca(finish(x)) → finish(a(x))
Cb(finish(x)) → finish(b(x))
Da(finish2(x)) → finish2(a(x))
Db(finish2(x)) → finish2(b(x))

The set Q consists of the following terms:

guess(a(x0))
guess(b(x0))
Ca(moveleft(Ba(x0)))
Cb(moveleft(Ba(x0)))
Ca(moveleft(Bb(x0)))
Cb(moveleft(Bb(x0)))
cut(moveleft(Ba(x0)))
cut(moveleft(Bb(x0)))
goright(Aa(x0))
goright(Ab(x0))
goright(wait(a(x0)))
goright(wait(b(x0)))
Ca(finish(x0))
Cb(finish(x0))
Da(finish2(x0))
Db(finish2(x0))

(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:

GUESS(a(x)) → CA(guess(x))
GUESS(a(x)) → GUESS(x)
GUESS(b(x)) → CB(guess(x))
GUESS(b(x)) → GUESS(x)
CUT(moveleft(Ba(x))) → DA(cut(goright(x)))
CUT(moveleft(Ba(x))) → CUT(goright(x))
CUT(moveleft(Ba(x))) → GORIGHT(x)
CUT(moveleft(Bb(x))) → DB(cut(goright(x)))
CUT(moveleft(Bb(x))) → CUT(goright(x))
CUT(moveleft(Bb(x))) → GORIGHT(x)
GORIGHT(Aa(x)) → CA(goright(x))
GORIGHT(Aa(x)) → GORIGHT(x)
GORIGHT(Ab(x)) → CB(goright(x))
GORIGHT(Ab(x)) → GORIGHT(x)

The TRS R consists of the following rules:

guess(a(x)) → Ca(guess(x))
guess(b(x)) → Cb(guess(x))
Ca(moveleft(Ba(x))) → moveleft(Ba(Aa(x)))
Cb(moveleft(Ba(x))) → moveleft(Ba(Ab(x)))
Ca(moveleft(Bb(x))) → moveleft(Bb(Aa(x)))
Cb(moveleft(Bb(x))) → moveleft(Bb(Ab(x)))
cut(moveleft(Ba(x))) → Da(cut(goright(x)))
cut(moveleft(Bb(x))) → Db(cut(goright(x)))
goright(Aa(x)) → Ca(goright(x))
goright(Ab(x)) → Cb(goright(x))
goright(wait(a(x))) → moveleft(Ba(wait(x)))
goright(wait(b(x))) → moveleft(Bb(wait(x)))
Ca(finish(x)) → finish(a(x))
Cb(finish(x)) → finish(b(x))
Da(finish2(x)) → finish2(a(x))
Db(finish2(x)) → finish2(b(x))

The set Q consists of the following terms:

guess(a(x0))
guess(b(x0))
Ca(moveleft(Ba(x0)))
Cb(moveleft(Ba(x0)))
Ca(moveleft(Bb(x0)))
Cb(moveleft(Bb(x0)))
cut(moveleft(Ba(x0)))
cut(moveleft(Bb(x0)))
goright(Aa(x0))
goright(Ab(x0))
goright(wait(a(x0)))
goright(wait(b(x0)))
Ca(finish(x0))
Cb(finish(x0))
Da(finish2(x0))
Db(finish2(x0))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 8 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

GORIGHT(Ab(x)) → GORIGHT(x)
GORIGHT(Aa(x)) → GORIGHT(x)

The TRS R consists of the following rules:

guess(a(x)) → Ca(guess(x))
guess(b(x)) → Cb(guess(x))
Ca(moveleft(Ba(x))) → moveleft(Ba(Aa(x)))
Cb(moveleft(Ba(x))) → moveleft(Ba(Ab(x)))
Ca(moveleft(Bb(x))) → moveleft(Bb(Aa(x)))
Cb(moveleft(Bb(x))) → moveleft(Bb(Ab(x)))
cut(moveleft(Ba(x))) → Da(cut(goright(x)))
cut(moveleft(Bb(x))) → Db(cut(goright(x)))
goright(Aa(x)) → Ca(goright(x))
goright(Ab(x)) → Cb(goright(x))
goright(wait(a(x))) → moveleft(Ba(wait(x)))
goright(wait(b(x))) → moveleft(Bb(wait(x)))
Ca(finish(x)) → finish(a(x))
Cb(finish(x)) → finish(b(x))
Da(finish2(x)) → finish2(a(x))
Db(finish2(x)) → finish2(b(x))

The set Q consists of the following terms:

guess(a(x0))
guess(b(x0))
Ca(moveleft(Ba(x0)))
Cb(moveleft(Ba(x0)))
Ca(moveleft(Bb(x0)))
Cb(moveleft(Bb(x0)))
cut(moveleft(Ba(x0)))
cut(moveleft(Bb(x0)))
goright(Aa(x0))
goright(Ab(x0))
goright(wait(a(x0)))
goright(wait(b(x0)))
Ca(finish(x0))
Cb(finish(x0))
Da(finish2(x0))
Db(finish2(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:

GORIGHT(Ab(x)) → GORIGHT(x)
GORIGHT(Aa(x)) → GORIGHT(x)

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

guess(a(x0))
guess(b(x0))
Ca(moveleft(Ba(x0)))
Cb(moveleft(Ba(x0)))
Ca(moveleft(Bb(x0)))
Cb(moveleft(Bb(x0)))
cut(moveleft(Ba(x0)))
cut(moveleft(Bb(x0)))
goright(Aa(x0))
goright(Ab(x0))
goright(wait(a(x0)))
goright(wait(b(x0)))
Ca(finish(x0))
Cb(finish(x0))
Da(finish2(x0))
Db(finish2(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].

guess(a(x0))
guess(b(x0))
Ca(moveleft(Ba(x0)))
Cb(moveleft(Ba(x0)))
Ca(moveleft(Bb(x0)))
Cb(moveleft(Bb(x0)))
cut(moveleft(Ba(x0)))
cut(moveleft(Bb(x0)))
goright(Aa(x0))
goright(Ab(x0))
goright(wait(a(x0)))
goright(wait(b(x0)))
Ca(finish(x0))
Cb(finish(x0))
Da(finish2(x0))
Db(finish2(x0))

(13) Obligation:

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

GORIGHT(Ab(x)) → GORIGHT(x)
GORIGHT(Aa(x)) → GORIGHT(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:

  • GORIGHT(Ab(x)) → GORIGHT(x)
    The graph contains the following edges 1 > 1

  • GORIGHT(Aa(x)) → GORIGHT(x)
    The graph contains the following edges 1 > 1

(15) YES

(16) Obligation:

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

CUT(moveleft(Bb(x))) → CUT(goright(x))
CUT(moveleft(Ba(x))) → CUT(goright(x))

The TRS R consists of the following rules:

guess(a(x)) → Ca(guess(x))
guess(b(x)) → Cb(guess(x))
Ca(moveleft(Ba(x))) → moveleft(Ba(Aa(x)))
Cb(moveleft(Ba(x))) → moveleft(Ba(Ab(x)))
Ca(moveleft(Bb(x))) → moveleft(Bb(Aa(x)))
Cb(moveleft(Bb(x))) → moveleft(Bb(Ab(x)))
cut(moveleft(Ba(x))) → Da(cut(goright(x)))
cut(moveleft(Bb(x))) → Db(cut(goright(x)))
goright(Aa(x)) → Ca(goright(x))
goright(Ab(x)) → Cb(goright(x))
goright(wait(a(x))) → moveleft(Ba(wait(x)))
goright(wait(b(x))) → moveleft(Bb(wait(x)))
Ca(finish(x)) → finish(a(x))
Cb(finish(x)) → finish(b(x))
Da(finish2(x)) → finish2(a(x))
Db(finish2(x)) → finish2(b(x))

The set Q consists of the following terms:

guess(a(x0))
guess(b(x0))
Ca(moveleft(Ba(x0)))
Cb(moveleft(Ba(x0)))
Ca(moveleft(Bb(x0)))
Cb(moveleft(Bb(x0)))
cut(moveleft(Ba(x0)))
cut(moveleft(Bb(x0)))
goright(Aa(x0))
goright(Ab(x0))
goright(wait(a(x0)))
goright(wait(b(x0)))
Ca(finish(x0))
Cb(finish(x0))
Da(finish2(x0))
Db(finish2(x0))

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

(17) 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.

(18) Obligation:

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

CUT(moveleft(Bb(x))) → CUT(goright(x))
CUT(moveleft(Ba(x))) → CUT(goright(x))

The TRS R consists of the following rules:

goright(Aa(x)) → Ca(goright(x))
goright(Ab(x)) → Cb(goright(x))
goright(wait(a(x))) → moveleft(Ba(wait(x)))
goright(wait(b(x))) → moveleft(Bb(wait(x)))
Cb(moveleft(Ba(x))) → moveleft(Ba(Ab(x)))
Cb(moveleft(Bb(x))) → moveleft(Bb(Ab(x)))
Cb(finish(x)) → finish(b(x))
Ca(moveleft(Ba(x))) → moveleft(Ba(Aa(x)))
Ca(moveleft(Bb(x))) → moveleft(Bb(Aa(x)))
Ca(finish(x)) → finish(a(x))

The set Q consists of the following terms:

guess(a(x0))
guess(b(x0))
Ca(moveleft(Ba(x0)))
Cb(moveleft(Ba(x0)))
Ca(moveleft(Bb(x0)))
Cb(moveleft(Bb(x0)))
cut(moveleft(Ba(x0)))
cut(moveleft(Bb(x0)))
goright(Aa(x0))
goright(Ab(x0))
goright(wait(a(x0)))
goright(wait(b(x0)))
Ca(finish(x0))
Cb(finish(x0))
Da(finish2(x0))
Db(finish2(x0))

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

(19) 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].

guess(a(x0))
guess(b(x0))
cut(moveleft(Ba(x0)))
cut(moveleft(Bb(x0)))
Da(finish2(x0))
Db(finish2(x0))

(20) Obligation:

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

CUT(moveleft(Bb(x))) → CUT(goright(x))
CUT(moveleft(Ba(x))) → CUT(goright(x))

The TRS R consists of the following rules:

goright(Aa(x)) → Ca(goright(x))
goright(Ab(x)) → Cb(goright(x))
goright(wait(a(x))) → moveleft(Ba(wait(x)))
goright(wait(b(x))) → moveleft(Bb(wait(x)))
Cb(moveleft(Ba(x))) → moveleft(Ba(Ab(x)))
Cb(moveleft(Bb(x))) → moveleft(Bb(Ab(x)))
Cb(finish(x)) → finish(b(x))
Ca(moveleft(Ba(x))) → moveleft(Ba(Aa(x)))
Ca(moveleft(Bb(x))) → moveleft(Bb(Aa(x)))
Ca(finish(x)) → finish(a(x))

The set Q consists of the following terms:

Ca(moveleft(Ba(x0)))
Cb(moveleft(Ba(x0)))
Ca(moveleft(Bb(x0)))
Cb(moveleft(Bb(x0)))
goright(Aa(x0))
goright(Ab(x0))
goright(wait(a(x0)))
goright(wait(b(x0)))
Ca(finish(x0))
Cb(finish(x0))

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

(21) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


CUT(moveleft(Bb(x))) → CUT(goright(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x1)) = 1 + x1   
POL(Ab(x1)) = 1 + x1   
POL(Ba(x1)) = x1   
POL(Bb(x1)) = 1 + x1   
POL(CUT(x1)) = x1   
POL(Ca(x1)) = 1 + x1   
POL(Cb(x1)) = 1 + x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = 1 + x1   
POL(finish(x1)) = 1 + x1   
POL(goright(x1)) = x1   
POL(moveleft(x1)) = x1   
POL(wait(x1)) = x1   

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

goright(Aa(x)) → Ca(goright(x))
goright(Ab(x)) → Cb(goright(x))
goright(wait(a(x))) → moveleft(Ba(wait(x)))
goright(wait(b(x))) → moveleft(Bb(wait(x)))
Cb(moveleft(Ba(x))) → moveleft(Ba(Ab(x)))
Cb(moveleft(Bb(x))) → moveleft(Bb(Ab(x)))
Cb(finish(x)) → finish(b(x))
Ca(moveleft(Ba(x))) → moveleft(Ba(Aa(x)))
Ca(moveleft(Bb(x))) → moveleft(Bb(Aa(x)))
Ca(finish(x)) → finish(a(x))

(22) Obligation:

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

CUT(moveleft(Ba(x))) → CUT(goright(x))

The TRS R consists of the following rules:

goright(Aa(x)) → Ca(goright(x))
goright(Ab(x)) → Cb(goright(x))
goright(wait(a(x))) → moveleft(Ba(wait(x)))
goright(wait(b(x))) → moveleft(Bb(wait(x)))
Cb(moveleft(Ba(x))) → moveleft(Ba(Ab(x)))
Cb(moveleft(Bb(x))) → moveleft(Bb(Ab(x)))
Cb(finish(x)) → finish(b(x))
Ca(moveleft(Ba(x))) → moveleft(Ba(Aa(x)))
Ca(moveleft(Bb(x))) → moveleft(Bb(Aa(x)))
Ca(finish(x)) → finish(a(x))

The set Q consists of the following terms:

Ca(moveleft(Ba(x0)))
Cb(moveleft(Ba(x0)))
Ca(moveleft(Bb(x0)))
Cb(moveleft(Bb(x0)))
goright(Aa(x0))
goright(Ab(x0))
goright(wait(a(x0)))
goright(wait(b(x0)))
Ca(finish(x0))
Cb(finish(x0))

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

(23) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


CUT(moveleft(Ba(x))) → CUT(goright(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x1)) = x1   
POL(Ab(x1)) = 1 + x1   
POL(Ba(x1)) = 1 + x1   
POL(Bb(x1)) = 0   
POL(CUT(x1)) = x1   
POL(Ca(x1)) = x1   
POL(Cb(x1)) = 1 + x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = x1   
POL(finish(x1)) = 1   
POL(goright(x1)) = x1   
POL(moveleft(x1)) = x1   
POL(wait(x1)) = x1   

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

goright(Aa(x)) → Ca(goright(x))
goright(Ab(x)) → Cb(goright(x))
goright(wait(a(x))) → moveleft(Ba(wait(x)))
goright(wait(b(x))) → moveleft(Bb(wait(x)))
Cb(moveleft(Ba(x))) → moveleft(Ba(Ab(x)))
Cb(moveleft(Bb(x))) → moveleft(Bb(Ab(x)))
Cb(finish(x)) → finish(b(x))
Ca(moveleft(Ba(x))) → moveleft(Ba(Aa(x)))
Ca(moveleft(Bb(x))) → moveleft(Bb(Aa(x)))
Ca(finish(x)) → finish(a(x))

(24) Obligation:

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

goright(Aa(x)) → Ca(goright(x))
goright(Ab(x)) → Cb(goright(x))
goright(wait(a(x))) → moveleft(Ba(wait(x)))
goright(wait(b(x))) → moveleft(Bb(wait(x)))
Cb(moveleft(Ba(x))) → moveleft(Ba(Ab(x)))
Cb(moveleft(Bb(x))) → moveleft(Bb(Ab(x)))
Cb(finish(x)) → finish(b(x))
Ca(moveleft(Ba(x))) → moveleft(Ba(Aa(x)))
Ca(moveleft(Bb(x))) → moveleft(Bb(Aa(x)))
Ca(finish(x)) → finish(a(x))

The set Q consists of the following terms:

Ca(moveleft(Ba(x0)))
Cb(moveleft(Ba(x0)))
Ca(moveleft(Bb(x0)))
Cb(moveleft(Bb(x0)))
goright(Aa(x0))
goright(Ab(x0))
goright(wait(a(x0)))
goright(wait(b(x0)))
Ca(finish(x0))
Cb(finish(x0))

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) YES

(27) Obligation:

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

GUESS(b(x)) → GUESS(x)
GUESS(a(x)) → GUESS(x)

The TRS R consists of the following rules:

guess(a(x)) → Ca(guess(x))
guess(b(x)) → Cb(guess(x))
Ca(moveleft(Ba(x))) → moveleft(Ba(Aa(x)))
Cb(moveleft(Ba(x))) → moveleft(Ba(Ab(x)))
Ca(moveleft(Bb(x))) → moveleft(Bb(Aa(x)))
Cb(moveleft(Bb(x))) → moveleft(Bb(Ab(x)))
cut(moveleft(Ba(x))) → Da(cut(goright(x)))
cut(moveleft(Bb(x))) → Db(cut(goright(x)))
goright(Aa(x)) → Ca(goright(x))
goright(Ab(x)) → Cb(goright(x))
goright(wait(a(x))) → moveleft(Ba(wait(x)))
goright(wait(b(x))) → moveleft(Bb(wait(x)))
Ca(finish(x)) → finish(a(x))
Cb(finish(x)) → finish(b(x))
Da(finish2(x)) → finish2(a(x))
Db(finish2(x)) → finish2(b(x))

The set Q consists of the following terms:

guess(a(x0))
guess(b(x0))
Ca(moveleft(Ba(x0)))
Cb(moveleft(Ba(x0)))
Ca(moveleft(Bb(x0)))
Cb(moveleft(Bb(x0)))
cut(moveleft(Ba(x0)))
cut(moveleft(Bb(x0)))
goright(Aa(x0))
goright(Ab(x0))
goright(wait(a(x0)))
goright(wait(b(x0)))
Ca(finish(x0))
Cb(finish(x0))
Da(finish2(x0))
Db(finish2(x0))

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

(28) 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.

(29) Obligation:

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

GUESS(b(x)) → GUESS(x)
GUESS(a(x)) → GUESS(x)

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

guess(a(x0))
guess(b(x0))
Ca(moveleft(Ba(x0)))
Cb(moveleft(Ba(x0)))
Ca(moveleft(Bb(x0)))
Cb(moveleft(Bb(x0)))
cut(moveleft(Ba(x0)))
cut(moveleft(Bb(x0)))
goright(Aa(x0))
goright(Ab(x0))
goright(wait(a(x0)))
goright(wait(b(x0)))
Ca(finish(x0))
Cb(finish(x0))
Da(finish2(x0))
Db(finish2(x0))

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

(30) 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].

guess(a(x0))
guess(b(x0))
Ca(moveleft(Ba(x0)))
Cb(moveleft(Ba(x0)))
Ca(moveleft(Bb(x0)))
Cb(moveleft(Bb(x0)))
cut(moveleft(Ba(x0)))
cut(moveleft(Bb(x0)))
goright(Aa(x0))
goright(Ab(x0))
goright(wait(a(x0)))
goright(wait(b(x0)))
Ca(finish(x0))
Cb(finish(x0))
Da(finish2(x0))
Db(finish2(x0))

(31) Obligation:

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

GUESS(b(x)) → GUESS(x)
GUESS(a(x)) → GUESS(x)

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

(32) 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:

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

  • GUESS(a(x)) → GUESS(x)
    The graph contains the following edges 1 > 1

(33) YES