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

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

0 QTRS
↳1 QTRSRRRProof (⇔, 40 ms)
↳2 QTRS
↳3 Overlay + Local Confluence (⇔, 2 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 QDPSizeChangeProof (⇔, 0 ms)
        ↳22 YES

(0) Obligation:

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

Begin(b(b(x))) → Wait(Right1(x))
Begin(b(x)) → Wait(Right2(x))
Right1(b(End(x))) → Left(a(End(x)))
Right2(b(b(End(x)))) → Left(a(End(x)))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
a(x) → b(b(x))
b(b(b(x))) → a(x)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = 9 + x1   
POL(Ab(x1)) = 4 + x1   
POL(Begin(x1)) = x1   
POL(End(x1)) = x1   
POL(Left(x1)) = x1   
POL(Right1(x1)) = 6 + x1   
POL(Right2(x1)) = 2 + x1   
POL(Wait(x1)) = 1 + x1   
POL(a(x1)) = 9 + x1   
POL(b(x1)) = 4 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

Begin(b(b(x))) → Wait(Right1(x))
Begin(b(x)) → Wait(Right2(x))
Right1(b(End(x))) → Left(a(End(x)))
Right2(b(b(End(x)))) → Left(a(End(x)))
Wait(Left(x)) → Begin(x)
a(x) → b(b(x))
b(b(b(x))) → a(x)


(2) Obligation:

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

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(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:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right1(b(x0))
Right2(b(x0))
Aa(Left(x0))
Ab(Left(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:

RIGHT1(a(x)) → AA(Right1(x))
RIGHT1(a(x)) → RIGHT1(x)
RIGHT2(a(x)) → AA(Right2(x))
RIGHT2(a(x)) → RIGHT2(x)
RIGHT1(b(x)) → AB(Right1(x))
RIGHT1(b(x)) → RIGHT1(x)
RIGHT2(b(x)) → AB(Right2(x))
RIGHT2(b(x)) → RIGHT2(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right1(b(x0))
Right2(b(x0))
Aa(Left(x0))
Ab(Left(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 2 SCCs with 4 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

RIGHT2(b(x)) → RIGHT2(x)
RIGHT2(a(x)) → RIGHT2(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right1(b(x0))
Right2(b(x0))
Aa(Left(x0))
Ab(Left(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:

RIGHT2(b(x)) → RIGHT2(x)
RIGHT2(a(x)) → RIGHT2(x)

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

Right1(a(x0))
Right2(a(x0))
Right1(b(x0))
Right2(b(x0))
Aa(Left(x0))
Ab(Left(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].

Right1(a(x0))
Right2(a(x0))
Right1(b(x0))
Right2(b(x0))
Aa(Left(x0))
Ab(Left(x0))

(13) Obligation:

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

RIGHT2(b(x)) → RIGHT2(x)
RIGHT2(a(x)) → RIGHT2(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:

  • RIGHT2(b(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(a(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

(15) YES

(16) Obligation:

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

RIGHT1(b(x)) → RIGHT1(x)
RIGHT1(a(x)) → RIGHT1(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right1(b(x0))
Right2(b(x0))
Aa(Left(x0))
Ab(Left(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:

RIGHT1(b(x)) → RIGHT1(x)
RIGHT1(a(x)) → RIGHT1(x)

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

Right1(a(x0))
Right2(a(x0))
Right1(b(x0))
Right2(b(x0))
Aa(Left(x0))
Ab(Left(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].

Right1(a(x0))
Right2(a(x0))
Right1(b(x0))
Right2(b(x0))
Aa(Left(x0))
Ab(Left(x0))

(20) Obligation:

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

RIGHT1(b(x)) → RIGHT1(x)
RIGHT1(a(x)) → RIGHT1(x)

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

(21) 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:

  • RIGHT1(b(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(a(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

(22) YES