YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z049-split.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 10 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 UsableRulesProof (⇔, 0 ms)
        ↳12 QDP
        ↳13 QDPSizeChangeProof (⇔, 0 ms)
        ↳14 YES
    ↳15 QDP
        ↳16 UsableRulesProof (⇔, 0 ms)
        ↳17 QDP
        ↳18 QDPSizeChangeProof (⇔, 0 ms)
        ↳19 YES
    ↳20 QDP
        ↳21 UsableRulesProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPSizeChangeProof (⇔, 0 ms)
        ↳24 YES
    ↳25 QDP
        ↳26 UsableRulesProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 QDPSizeChangeProof (⇔, 0 ms)
        ↳29 YES
    ↳30 QDP
        ↳31 UsableRulesProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 QDPSizeChangeProof (⇔, 0 ms)
        ↳34 YES
    ↳35 QDP
        ↳36 UsableRulesProof (⇔, 0 ms)
        ↳37 QDP
        ↳38 QDPOrderProof (⇔, 2501 ms)
        ↳39 QDP
        ↳40 UsableRulesProof (⇔, 0 ms)
        ↳41 QDP
        ↳42 QDPOrderProof (⇔, 716 ms)
        ↳43 QDP
        ↳44 UsableRulesProof (⇔, 0 ms)
        ↳45 QDP
        ↳46 QDPOrderProof (⇔, 1247 ms)
        ↳47 QDP
        ↳48 UsableRulesProof (⇔, 0 ms)
        ↳49 QDP
        ↳50 QDPOrderProof (⇔, 777 ms)
        ↳51 QDP
        ↳52 DependencyGraphProof (⇔, 0 ms)
        ↳53 TRUE

(0) Obligation:

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

Begin(b(a(a(b(b(x)))))) → Wait(Right1(x))
Begin(a(a(b(b(x))))) → Wait(Right2(x))
Begin(a(b(b(x)))) → Wait(Right3(x))
Begin(b(b(x))) → Wait(Right4(x))
Begin(b(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right4(a(b(a(a(End(x)))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right5(a(b(a(a(b(End(x))))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))

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:

BEGIN(b(a(a(b(b(x)))))) → WAIT(Right1(x))
BEGIN(b(a(a(b(b(x)))))) → RIGHT1(x)
BEGIN(a(a(b(b(x))))) → WAIT(Right2(x))
BEGIN(a(a(b(b(x))))) → RIGHT2(x)
BEGIN(a(b(b(x)))) → WAIT(Right3(x))
BEGIN(a(b(b(x)))) → RIGHT3(x)
BEGIN(b(b(x))) → WAIT(Right4(x))
BEGIN(b(b(x))) → RIGHT4(x)
BEGIN(b(x)) → WAIT(Right5(x))
BEGIN(b(x)) → RIGHT5(x)
RIGHT1(a(End(x))) → A(a(b(a(b(b(a(End(x))))))))
RIGHT1(a(End(x))) → A(b(a(b(b(a(End(x)))))))
RIGHT1(a(End(x))) → A(b(b(a(End(x)))))
RIGHT2(a(b(End(x)))) → A(a(b(a(b(b(a(End(x))))))))
RIGHT2(a(b(End(x)))) → A(b(a(b(b(a(End(x)))))))
RIGHT2(a(b(End(x)))) → A(b(b(a(End(x)))))
RIGHT2(a(b(End(x)))) → A(End(x))
RIGHT3(a(b(a(End(x))))) → A(a(b(a(b(b(a(End(x))))))))
RIGHT3(a(b(a(End(x))))) → A(b(a(b(b(a(End(x)))))))
RIGHT3(a(b(a(End(x))))) → A(b(b(a(End(x)))))
RIGHT4(a(b(a(a(End(x)))))) → A(a(b(a(b(b(a(End(x))))))))
RIGHT4(a(b(a(a(End(x)))))) → A(b(a(b(b(a(End(x)))))))
RIGHT4(a(b(a(a(End(x)))))) → A(b(b(a(End(x)))))
RIGHT5(a(b(a(a(b(End(x))))))) → A(a(b(a(b(b(a(End(x))))))))
RIGHT5(a(b(a(a(b(End(x))))))) → A(b(a(b(b(a(End(x)))))))
RIGHT5(a(b(a(a(b(End(x))))))) → A(b(b(a(End(x)))))
RIGHT5(a(b(a(a(b(End(x))))))) → A(End(x))
RIGHT1(a(x)) → AA(Right1(x))
RIGHT1(a(x)) → RIGHT1(x)
RIGHT2(a(x)) → AA(Right2(x))
RIGHT2(a(x)) → RIGHT2(x)
RIGHT3(a(x)) → AA(Right3(x))
RIGHT3(a(x)) → RIGHT3(x)
RIGHT4(a(x)) → AA(Right4(x))
RIGHT4(a(x)) → RIGHT4(x)
RIGHT5(a(x)) → AA(Right5(x))
RIGHT5(a(x)) → RIGHT5(x)
RIGHT1(b(x)) → AB(Right1(x))
RIGHT1(b(x)) → RIGHT1(x)
RIGHT2(b(x)) → AB(Right2(x))
RIGHT2(b(x)) → RIGHT2(x)
RIGHT3(b(x)) → AB(Right3(x))
RIGHT3(b(x)) → RIGHT3(x)
RIGHT4(b(x)) → AB(Right4(x))
RIGHT4(b(x)) → RIGHT4(x)
RIGHT5(b(x)) → AB(Right5(x))
RIGHT5(b(x)) → RIGHT5(x)
AA(Left(x)) → A(x)
WAIT(Left(x)) → BEGIN(x)
A(b(a(a(b(b(x)))))) → A(a(b(a(b(b(a(x)))))))
A(b(a(a(b(b(x)))))) → A(b(a(b(b(a(x))))))
A(b(a(a(b(b(x)))))) → A(b(b(a(x))))
A(b(a(a(b(b(x)))))) → A(x)

The TRS R consists of the following rules:

Begin(b(a(a(b(b(x)))))) → Wait(Right1(x))
Begin(a(a(b(b(x))))) → Wait(Right2(x))
Begin(a(b(b(x)))) → Wait(Right3(x))
Begin(b(b(x))) → Wait(Right4(x))
Begin(b(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right4(a(b(a(a(End(x)))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right5(a(b(a(a(b(End(x))))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))

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 7 SCCs with 36 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

A(b(a(a(b(b(x)))))) → A(x)

The TRS R consists of the following rules:

Begin(b(a(a(b(b(x)))))) → Wait(Right1(x))
Begin(a(a(b(b(x))))) → Wait(Right2(x))
Begin(a(b(b(x)))) → Wait(Right3(x))
Begin(b(b(x))) → Wait(Right4(x))
Begin(b(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right4(a(b(a(a(End(x)))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right5(a(b(a(a(b(End(x))))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))

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:

A(b(a(a(b(b(x)))))) → A(x)

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:

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

(9) YES

(10) Obligation:

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

RIGHT5(b(x)) → RIGHT5(x)
RIGHT5(a(x)) → RIGHT5(x)

The TRS R consists of the following rules:

Begin(b(a(a(b(b(x)))))) → Wait(Right1(x))
Begin(a(a(b(b(x))))) → Wait(Right2(x))
Begin(a(b(b(x)))) → Wait(Right3(x))
Begin(b(b(x))) → Wait(Right4(x))
Begin(b(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right4(a(b(a(a(End(x)))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right5(a(b(a(a(b(End(x))))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))

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

(11) 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.

(12) Obligation:

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

RIGHT5(b(x)) → RIGHT5(x)
RIGHT5(a(x)) → RIGHT5(x)

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

(13) 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:

  • RIGHT5(b(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

  • RIGHT5(a(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

(14) YES

(15) Obligation:

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

RIGHT4(b(x)) → RIGHT4(x)
RIGHT4(a(x)) → RIGHT4(x)

The TRS R consists of the following rules:

Begin(b(a(a(b(b(x)))))) → Wait(Right1(x))
Begin(a(a(b(b(x))))) → Wait(Right2(x))
Begin(a(b(b(x)))) → Wait(Right3(x))
Begin(b(b(x))) → Wait(Right4(x))
Begin(b(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right4(a(b(a(a(End(x)))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right5(a(b(a(a(b(End(x))))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))

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

(16) 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.

(17) Obligation:

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

RIGHT4(b(x)) → RIGHT4(x)
RIGHT4(a(x)) → RIGHT4(x)

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

(18) 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:

  • RIGHT4(b(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(a(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

(19) YES

(20) Obligation:

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

RIGHT3(b(x)) → RIGHT3(x)
RIGHT3(a(x)) → RIGHT3(x)

The TRS R consists of the following rules:

Begin(b(a(a(b(b(x)))))) → Wait(Right1(x))
Begin(a(a(b(b(x))))) → Wait(Right2(x))
Begin(a(b(b(x)))) → Wait(Right3(x))
Begin(b(b(x))) → Wait(Right4(x))
Begin(b(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right4(a(b(a(a(End(x)))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right5(a(b(a(a(b(End(x))))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))

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:

RIGHT3(b(x)) → RIGHT3(x)
RIGHT3(a(x)) → RIGHT3(x)

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

(23) 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:

  • RIGHT3(b(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

  • RIGHT3(a(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

(24) YES

(25) 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:

Begin(b(a(a(b(b(x)))))) → Wait(Right1(x))
Begin(a(a(b(b(x))))) → Wait(Right2(x))
Begin(a(b(b(x)))) → Wait(Right3(x))
Begin(b(b(x))) → Wait(Right4(x))
Begin(b(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right4(a(b(a(a(End(x)))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right5(a(b(a(a(b(End(x))))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))

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

(26) 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.

(27) 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.

(28) 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

(29) YES

(30) 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:

Begin(b(a(a(b(b(x)))))) → Wait(Right1(x))
Begin(a(a(b(b(x))))) → Wait(Right2(x))
Begin(a(b(b(x)))) → Wait(Right3(x))
Begin(b(b(x))) → Wait(Right4(x))
Begin(b(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right4(a(b(a(a(End(x)))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right5(a(b(a(a(b(End(x))))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))

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

(31) 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.

(32) 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.

(33) 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

(34) YES

(35) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(b(a(a(b(b(x)))))) → WAIT(Right1(x))
BEGIN(a(a(b(b(x))))) → WAIT(Right2(x))
BEGIN(a(b(b(x)))) → WAIT(Right3(x))
BEGIN(b(b(x))) → WAIT(Right4(x))
BEGIN(b(x)) → WAIT(Right5(x))

The TRS R consists of the following rules:

Begin(b(a(a(b(b(x)))))) → Wait(Right1(x))
Begin(a(a(b(b(x))))) → Wait(Right2(x))
Begin(a(b(b(x)))) → Wait(Right3(x))
Begin(b(b(x))) → Wait(Right4(x))
Begin(b(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right4(a(b(a(a(End(x)))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right5(a(b(a(a(b(End(x))))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))

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

(36) 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.

(37) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(b(a(a(b(b(x)))))) → WAIT(Right1(x))
BEGIN(a(a(b(b(x))))) → WAIT(Right2(x))
BEGIN(a(b(b(x)))) → WAIT(Right3(x))
BEGIN(b(b(x))) → WAIT(Right4(x))
BEGIN(b(x)) → WAIT(Right5(x))

The TRS R consists of the following rules:

Right5(a(b(a(a(b(End(x))))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right5(a(x)) → Aa(Right5(x))
Right5(b(x)) → Ab(Right5(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))
Right4(a(b(a(a(End(x)))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right4(a(x)) → Aa(Right4(x))
Right4(b(x)) → Ab(Right4(x))
Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(x)) → Aa(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(x)) → Aa(Right2(x))
Right2(b(x)) → Ab(Right2(x))
Right1(a(End(x))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right1(a(x)) → Aa(Right1(x))
Right1(b(x)) → Ab(Right1(x))

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

(38) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


BEGIN(b(a(a(b(b(x)))))) → WAIT(Right1(x))
BEGIN(b(b(x))) → WAIT(Right4(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(WAIT(x1)) = -I +
[0A,-I,0A]
·x1

POL(Left(x1)) =
/0A\
|0A|
\-I/
 +
/-I-I0A\
|0A0A0A|
\0A-I0A/
·x1

POL(BEGIN(x1)) = 0A +
[0A,-I,0A]
·x1

POL(b(x1)) =
/1A\
|0A|
\0A/
 +
/1A0A0A\
|-I0A0A|
\0A-I0A/
·x1

POL(a(x1)) =
/0A\
|0A|
\0A/
 +
/-I-I-I\
|0A0A-I|
\-I-I0A/
·x1

POL(Right1(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A0A\
|1A1A1A|
\1A0A0A/
·x1

POL(Right2(x1)) =
/-I\
|-I|
\1A/
 +
/0A0A0A\
|1A1A0A|
\1A0A0A/
·x1

POL(Right3(x1)) =
/0A\
|0A|
\-I/
 +
/0A0A0A\
|1A1A0A|
\1A0A0A/
·x1

POL(Right4(x1)) =
/0A\
|0A|
\-I/
 +
/0A0A0A\
|1A1A0A|
\1A0A0A/
·x1

POL(Right5(x1)) =
/-I\
|-I|
\-I/
 +
/0A0A0A\
|1A1A0A|
\1A0A0A/
·x1

POL(End(x1)) =
/0A\
|0A|
\1A/
 +
/-I-I-I\
|0A-I-I|
\0A-I0A/
·x1

POL(Aa(x1)) =
/0A\
|0A|
\0A/
 +
/0A-I-I\
|0A0A0A|
\0A-I-I/
·x1

POL(Ab(x1)) =
/1A\
|1A|
\1A/
 +
/0A-I0A\
|0A0A1A|
\0A0A1A/
·x1

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

Right1(a(End(x))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right1(a(x)) → Aa(Right1(x))
Right1(b(x)) → Ab(Right1(x))
Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(x)) → Aa(Right2(x))
Right2(b(x)) → Ab(Right2(x))
Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(x)) → Aa(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Right4(a(b(a(a(End(x)))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right4(a(x)) → Aa(Right4(x))
Right4(b(x)) → Ab(Right4(x))
Right5(a(b(a(a(b(End(x))))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right5(a(x)) → Aa(Right5(x))
Right5(b(x)) → Ab(Right5(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))

(39) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(a(b(b(x))))) → WAIT(Right2(x))
BEGIN(a(b(b(x)))) → WAIT(Right3(x))
BEGIN(b(x)) → WAIT(Right5(x))

The TRS R consists of the following rules:

Right5(a(b(a(a(b(End(x))))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right5(a(x)) → Aa(Right5(x))
Right5(b(x)) → Ab(Right5(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))
Right4(a(b(a(a(End(x)))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right4(a(x)) → Aa(Right4(x))
Right4(b(x)) → Ab(Right4(x))
Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(x)) → Aa(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(x)) → Aa(Right2(x))
Right2(b(x)) → Ab(Right2(x))
Right1(a(End(x))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right1(a(x)) → Aa(Right1(x))
Right1(b(x)) → Ab(Right1(x))

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

(40) 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.

(41) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(a(b(b(x))))) → WAIT(Right2(x))
BEGIN(a(b(b(x)))) → WAIT(Right3(x))
BEGIN(b(x)) → WAIT(Right5(x))

The TRS R consists of the following rules:

Right5(a(b(a(a(b(End(x))))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right5(a(x)) → Aa(Right5(x))
Right5(b(x)) → Ab(Right5(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))
Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(x)) → Aa(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(x)) → Aa(Right2(x))
Right2(b(x)) → Ab(Right2(x))

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

(42) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


BEGIN(b(x)) → WAIT(Right5(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(WAIT(x1)) = -I +
[0A,-I,0A]
·x1

POL(Left(x1)) =
/0A\
|1A|
\0A/
 +
/1A1A1A\
|1A1A1A|
\0A0A0A/
·x1

POL(BEGIN(x1)) = 0A +
[-I,1A,1A]
·x1

POL(a(x1)) =
/0A\
|-I|
\-I/
 +
/-I0A-I\
|-I-I0A|
\-I0A0A/
·x1

POL(b(x1)) =
/-I\
|-I|
\-I/
 +
/0A-I-I\
|0A-I-I|
\0A0A0A/
·x1

POL(Right2(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A1A\
|-I0A1A|
\-I-I1A/
·x1

POL(Right3(x1)) =
/0A\
|1A|
\0A/
 +
/1A0A0A\
|-I0A0A|
\1A0A0A/
·x1

POL(Right5(x1)) =
/-I\
|0A|
\-I/
 +
/0A0A0A\
|-I0A0A|
\0A0A0A/
·x1

POL(End(x1)) =
/1A\
|0A|
\0A/
 +
/0A0A-I\
|-I-I-I|
\-I-I-I/
·x1

POL(Aa(x1)) =
/0A\
|0A|
\0A/
 +
/-I0A-I\
|-I0A-I|
\-I0A-I/
·x1

POL(Ab(x1)) =
/-I\
|0A|
\-I/
 +
/0A-I0A\
|-I0A-I|
\0A-I0A/
·x1

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

Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(x)) → Aa(Right2(x))
Right2(b(x)) → Ab(Right2(x))
Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(x)) → Aa(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Right5(a(b(a(a(b(End(x))))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right5(a(x)) → Aa(Right5(x))
Right5(b(x)) → Ab(Right5(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))

(43) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(a(b(b(x))))) → WAIT(Right2(x))
BEGIN(a(b(b(x)))) → WAIT(Right3(x))

The TRS R consists of the following rules:

Right5(a(b(a(a(b(End(x))))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right5(a(x)) → Aa(Right5(x))
Right5(b(x)) → Ab(Right5(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))
Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(x)) → Aa(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(x)) → Aa(Right2(x))
Right2(b(x)) → Ab(Right2(x))

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

(44) 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.

(45) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(a(b(b(x))))) → WAIT(Right2(x))
BEGIN(a(b(b(x)))) → WAIT(Right3(x))

The TRS R consists of the following rules:

Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(x)) → Aa(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))
Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(x)) → Aa(Right2(x))
Right2(b(x)) → Ab(Right2(x))

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

(46) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


BEGIN(a(a(b(b(x))))) → WAIT(Right2(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(WAIT(x1)) = 0A +
[0A,0A,0A]
·x1

POL(Left(x1)) =
/1A\
|-I|
\0A/
 +
/1A1A0A\
|0A0A0A|
\0A0A1A/
·x1

POL(BEGIN(x1)) = 0A +
[1A,0A,0A]
·x1

POL(a(x1)) =
/-I\
|-I|
\-I/
 +
/0A0A-I\
|0A0A-I|
\-I0A-I/
·x1

POL(b(x1)) =
/0A\
|-I|
\-I/
 +
/0A0A0A\
|-I0A0A|
\0A0A0A/
·x1

POL(Right2(x1)) =
/0A\
|0A|
\0A/
 +
/-I0A-I\
|0A0A-I|
\-I-I0A/
·x1

POL(Right3(x1)) =
/0A\
|0A|
\1A/
 +
/1A0A-I\
|1A0A-I|
\1A0A0A/
·x1

POL(End(x1)) =
/-I\
|-I|
\1A/
 +
/0A0A0A\
|-I0A0A|
\1A1A1A/
·x1

POL(Aa(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A-I\
|0A0A-I|
\0A-I-I/
·x1

POL(Ab(x1)) =
/0A\
|0A|
\0A/
 +
/0A-I0A\
|0A0A-I|
\0A0A0A/
·x1

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

Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(x)) → Aa(Right2(x))
Right2(b(x)) → Ab(Right2(x))
Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(x)) → Aa(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))

(47) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(b(b(x)))) → WAIT(Right3(x))

The TRS R consists of the following rules:

Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(x)) → Aa(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))
Right2(a(b(End(x)))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right2(a(x)) → Aa(Right2(x))
Right2(b(x)) → Ab(Right2(x))

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

(48) 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.

(49) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(b(b(x)))) → WAIT(Right3(x))

The TRS R consists of the following rules:

Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(x)) → Aa(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))

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

(50) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


WAIT(Left(x)) → BEGIN(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(WAIT(x1)) = -I +
[0A,0A,0A]
·x1

POL(Left(x1)) =
/0A\
|0A|
\1A/
 +
/-I0A0A\
|0A-I0A|
\0A1A0A/
·x1

POL(BEGIN(x1)) = 0A +
[-I,0A,-I]
·x1

POL(a(x1)) =
/0A\
|-I|
\0A/
 +
/-I-I-I\
|0A-I0A|
\-I-I-I/
·x1

POL(b(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A0A\
|0A1A0A|
\0A0A0A/
·x1

POL(Right3(x1)) =
/0A\
|-I|
\0A/
 +
/0A0A0A\
|0A-I0A|
\0A1A0A/
·x1

POL(End(x1)) =
/-I\
|0A|
\-I/
 +
/0A0A-I\
|0A0A0A|
\-I0A-I/
·x1

POL(Aa(x1)) =
/0A\
|0A|
\0A/
 +
/-I0A-I\
|-I-I-I|
\-I1A-I/
·x1

POL(Ab(x1)) =
/-I\
|-I|
\-I/
 +
/0A0A0A\
|0A0A-I|
\0A0A1A/
·x1

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

Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(x)) → Aa(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))

(51) Obligation:

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

BEGIN(a(b(b(x)))) → WAIT(Right3(x))

The TRS R consists of the following rules:

Right3(a(b(a(End(x))))) → Left(a(a(b(a(b(b(a(End(x)))))))))
Right3(a(x)) → Aa(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
a(b(a(a(b(b(x)))))) → a(a(b(a(b(b(a(x)))))))

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

(52) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(53) TRUE