Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema

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

0 QTRS

↳1 DependencyPairsProof (⇔, 51 ms)

↳2 QDP

↳3 DependencyGraphProof (⇔, 0 ms)

↳4 AND

    ↳5 QDP

        ↳6 UsableRulesProof (⇔, 0 ms)

        ↳7 QDP

        ↳8 QDPOrderProof (⇔, 90 ms)

        ↳9 QDP

        ↳10 QDPOrderProof (⇔, 1158 ms)

        ↳11 QDP

        ↳12 PisEmptyProof (⇔, 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 QDPSizeChangeProof (⇔, 0 ms)

        ↳33 YES

    ↳34 QDP

        ↳35 UsableRulesProof (⇔, 0 ms)

        ↳36 QDP

        ↳37 QDPSizeChangeProof (⇔, 0 ms)

        ↳38 YES

    ↳39 QDP

        ↳40 UsableRulesProof (⇔, 0 ms)

        ↳41 QDP

        ↳42 QDPOrderProof (⇔, 4830 ms)

        ↳43 QDP

        ↳44 UsableRulesProof (⇔, 0 ms)

        ↳45 QDP

        ↳46 QDPOrderProof (⇔, 1608 ms)

        ↳47 QDP

        ↳48 UsableRulesProof (⇔, 0 ms)

        ↳49 QDP

        ↳50 QDPOrderProof (⇔, 760 ms)

        ↳51 QDP

        ↳52 UsableRulesProof (⇔, 0 ms)

        ↳53 QDP

        ↳54 QDPOrderProof (⇔, 8 ms)

        ↳55 QDP

        ↳56 DependencyGraphProof (⇔, 0 ms)

        ↳57 TRUE

(0) Obligation:

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

Begin(a(b(b(b(b(x)))))) → Wait(Right1(x))
Begin(b(b(b(b(x))))) → Wait(Right2(x))
Begin(b(b(b(x)))) → Wait(Right3(x))
Begin(b(b(x))) → Wait(Right4(x))
Begin(b(x)) → Wait(Right5(x))
Right1(b(End(x))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right2(b(a(End(x)))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right3(b(a(b(End(x))))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right4(b(a(b(b(End(x)))))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right5(b(a(b(b(b(End(x))))))) → Left(a(b(b(b(b(b(a(End(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))
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))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
Wait(Left(x)) → Begin(x)
b(a(b(b(b(b(x)))))) → a(b(b(b(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(a(b(b(b(b(x)))))) → WAIT(Right1(x))
BEGIN(a(b(b(b(b(x)))))) → RIGHT1(x)
BEGIN(b(b(b(b(x))))) → WAIT(Right2(x))
BEGIN(b(b(b(b(x))))) → RIGHT2(x)
BEGIN(b(b(b(x)))) → WAIT(Right3(x))
BEGIN(b(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(b(End(x))) → B(b(b(b(b(a(End(x)))))))
RIGHT1(b(End(x))) → B(b(b(b(a(End(x))))))
RIGHT1(b(End(x))) → B(b(b(a(End(x)))))
RIGHT1(b(End(x))) → B(b(a(End(x))))
RIGHT1(b(End(x))) → B(a(End(x)))
RIGHT2(b(a(End(x)))) → B(b(b(b(b(a(End(x)))))))
RIGHT2(b(a(End(x)))) → B(b(b(b(a(End(x))))))
RIGHT2(b(a(End(x)))) → B(b(b(a(End(x)))))
RIGHT2(b(a(End(x)))) → B(b(a(End(x))))
RIGHT3(b(a(b(End(x))))) → B(b(b(b(b(a(End(x)))))))
RIGHT3(b(a(b(End(x))))) → B(b(b(b(a(End(x))))))
RIGHT3(b(a(b(End(x))))) → B(b(b(a(End(x)))))
RIGHT3(b(a(b(End(x))))) → B(b(a(End(x))))
RIGHT3(b(a(b(End(x))))) → B(a(End(x)))
RIGHT4(b(a(b(b(End(x)))))) → B(b(b(b(b(a(End(x)))))))
RIGHT4(b(a(b(b(End(x)))))) → B(b(b(b(a(End(x))))))
RIGHT4(b(a(b(b(End(x)))))) → B(b(b(a(End(x)))))
RIGHT4(b(a(b(b(End(x)))))) → B(b(a(End(x))))
RIGHT4(b(a(b(b(End(x)))))) → B(a(End(x)))
RIGHT5(b(a(b(b(b(End(x))))))) → B(b(b(b(b(a(End(x)))))))
RIGHT5(b(a(b(b(b(End(x))))))) → B(b(b(b(a(End(x))))))
RIGHT5(b(a(b(b(b(End(x))))))) → B(b(b(a(End(x)))))
RIGHT5(b(a(b(b(b(End(x))))))) → B(b(a(End(x))))
RIGHT5(b(a(b(b(b(End(x))))))) → B(a(End(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)
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)
AB(Left(x)) → B(x)
WAIT(Left(x)) → BEGIN(x)
B(a(b(b(b(b(x)))))) → B(b(b(b(b(a(x))))))
B(a(b(b(b(b(x)))))) → B(b(b(b(a(x)))))
B(a(b(b(b(b(x)))))) → B(b(b(a(x))))
B(a(b(b(b(b(x)))))) → B(b(a(x)))
B(a(b(b(b(b(x)))))) → B(a(x))

The TRS R consists of the following rules:

Begin(a(b(b(b(b(x)))))) → Wait(Right1(x))
Begin(b(b(b(b(x))))) → Wait(Right2(x))
Begin(b(b(b(x)))) → Wait(Right3(x))
Begin(b(b(x))) → Wait(Right4(x))
Begin(b(x)) → Wait(Right5(x))
Right1(b(End(x))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right2(b(a(End(x)))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right3(b(a(b(End(x))))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right4(b(a(b(b(End(x)))))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right5(b(a(b(b(b(End(x))))))) → Left(a(b(b(b(b(b(a(End(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))
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))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
Wait(Left(x)) → Begin(x)
b(a(b(b(b(b(x)))))) → a(b(b(b(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 40 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

B(a(b(b(b(b(x)))))) → B(b(b(b(a(x)))))
B(a(b(b(b(b(x)))))) → B(b(b(b(b(a(x))))))
B(a(b(b(b(b(x)))))) → B(b(b(a(x))))
B(a(b(b(b(b(x)))))) → B(b(a(x)))
B(a(b(b(b(b(x)))))) → B(a(x))

The TRS R consists of the following rules:

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

B(a(b(b(b(b(x)))))) → B(b(b(b(a(x)))))
B(a(b(b(b(b(x)))))) → B(b(b(b(b(a(x))))))
B(a(b(b(b(b(x)))))) → B(b(b(a(x))))
B(a(b(b(b(b(x)))))) → B(b(a(x)))
B(a(b(b(b(b(x)))))) → B(a(x))

The TRS R consists of the following rules:

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

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

(8) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

B(a(b(b(b(b(x)))))) → B(b(b(b(a(x)))))
B(a(b(b(b(b(x)))))) → B(b(b(a(x))))
B(a(b(b(b(b(x)))))) → B(b(a(x)))
B(a(b(b(b(b(x)))))) → B(a(x))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(B(x₁)) = x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = 1 + x₁

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

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

(9) Obligation:

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

B(a(b(b(b(b(x)))))) → B(b(b(b(b(a(x))))))

The TRS R consists of the following rules:

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

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

(10) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

B(a(b(b(b(b(x)))))) → B(b(b(b(b(a(x))))))

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(B(x₁)) = -I +
[ 0A, 0A, 0A ]

· x₁

POL(a(x₁)) =
/ 0A \

| 0A |

\ -I /

+
/ -I -I 0A \

| 0A -I 1A |

\ -I -I -I /

· x₁

POL(b(x₁)) =
/ 1A \

| 0A |

\ 0A /

+
/ 0A 0A 1A \

| -I 0A 0A |

\ 0A -I 0A /

· x₁

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

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

(11) Obligation:

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

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

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

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) YES

(14) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(15) UsableRulesProof (EQUIVALENT transformation)

(16) Obligation:

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

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

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

(18) YES

(19) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(20) UsableRulesProof (EQUIVALENT transformation)

(21) Obligation:

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

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

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

(23) YES

(24) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(25) UsableRulesProof (EQUIVALENT transformation)

(26) Obligation:

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

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

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

(28) YES

(29) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(30) UsableRulesProof (EQUIVALENT transformation)

(31) Obligation:

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

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

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

(33) YES

(34) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(35) UsableRulesProof (EQUIVALENT transformation)

(36) Obligation:

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

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

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

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

(38) YES

(39) Obligation:

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

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

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

(40) UsableRulesProof (EQUIVALENT transformation)

(41) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(b(b(b(b(x)))))) → WAIT(Right1(x))
BEGIN(b(b(b(b(x))))) → WAIT(Right2(x))
BEGIN(b(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(b(a(b(b(b(End(x))))))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right5(b(x)) → Ab(Right5(x))
Right5(a(x)) → Aa(Right5(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
b(a(b(b(b(b(x)))))) → a(b(b(b(b(b(a(x)))))))
Right4(b(a(b(b(End(x)))))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right3(b(a(b(End(x))))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right2(b(a(End(x)))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right2(b(x)) → Ab(Right2(x))
Right2(a(x)) → Aa(Right2(x))
Right1(b(End(x))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right1(b(x)) → Ab(Right1(x))
Right1(a(x)) → Aa(Right1(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, 0A, 0A ]

· x₁

POL(Left(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ -I 1A 0A \

| 1A 0A 1A |

\ -I -I -I /

· x₁

POL(BEGIN(x₁)) = 0A +
[ 1A, 0A, 0A ]

· x₁

POL(a(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A -I -I \

| 0A -I -I |

\ 0A -I -I /

· x₁

POL(b(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A 0A 0A \

| 0A 0A 0A |

\ 0A 0A 0A /

· x₁

POL(Right1(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A -I -I \

| 0A -I -I |

\ 0A -I -I /

· x₁

POL(Right2(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 1A -I -I \

| 1A -I -I |

\ 1A -I -I /

· x₁

POL(Right3(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A -I -I \

| 0A -I -I |

\ 0A -I -I /

· x₁

POL(Right4(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A -I -I \

| 0A -I -I |

\ 0A -I -I /

· x₁

POL(Right5(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A -I -I \

| 0A -I -I |

\ 0A -I -I /

· x₁

POL(End(x₁)) =
/ -I \

| 0A |

\ 0A /

+
/ 0A 0A 0A \

| 1A 0A 1A |

\ 0A 1A 0A /

· x₁

POL(Ab(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A 0A 0A \

| 0A 0A 0A |

\ -I -I 0A /

· x₁

POL(Aa(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A 0A 0A \

| 0A 0A 0A |

\ 0A -I 0A /

· x₁

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

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

(43) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(44) UsableRulesProof (EQUIVALENT transformation)

(45) Obligation:

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

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

The TRS R consists of the following rules:

Right4(b(a(b(b(End(x)))))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
b(a(b(b(b(b(x)))))) → a(b(b(b(b(b(a(x)))))))
Right3(b(a(b(End(x))))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right2(b(a(End(x)))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right2(b(x)) → Ab(Right2(x))
Right2(a(x)) → Aa(Right2(x))
Right1(b(End(x))) → Left(a(b(b(b(b(b(a(End(x)))))))))
Right1(b(x)) → Ab(Right1(x))
Right1(a(x)) → Aa(Right1(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(b(b(b(x)))) → WAIT(Right3(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)) =
/ 0A \

| 1A |

\ 0A /

+
/ 0A 0A 0A \

| 1A 1A 1A |

\ 0A 0A 0A /

· x₁

POL(BEGIN(x₁)) = 0A +
[ 1A, 0A, 0A ]

· x₁

POL(a(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A 0A -I \

| 0A -I -I |

\ 0A -I -I /

· x₁

POL(b(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A 0A -I \

| -I 0A 0A |

\ -I 0A 0A /

· x₁

POL(Right1(x₁)) =
/ 0A \

| 1A |

\ 1A /

+
/ 0A 0A -I \

| 0A 0A -I |

\ 0A 0A -I /

· x₁

POL(Right2(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A 0A -I \

| 1A 0A -I |

\ 1A 0A -I /

· x₁

POL(Right3(x₁)) =
/ -I \

| 0A |

\ 0A /

+
/ 0A 0A -I \

| 0A 0A -I |

\ 0A 0A -I /

· x₁

POL(Right4(x₁)) =
/ 0A \

| 1A |

\ 1A /

+
/ 0A 0A -I \

| 0A 0A -I |

\ 0A 0A -I /

· x₁

POL(End(x₁)) =
/ 0A \

| 0A |

\ 1A /

+
/ -I 0A 0A \

| 0A 0A 0A |

\ 1A 1A 1A /

· x₁

POL(Ab(x₁)) =
/ 0A \

| -I |

\ 0A /

+
/ 0A -I -I \

| 0A 0A 0A |

\ 0A 0A 0A /

· x₁

POL(Aa(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A -I -I \

| 0A 0A 0A |

\ 0A 0A 0A /

· x₁

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

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

The TRS R consists of the following rules:

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

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

(48) UsableRulesProof (EQUIVALENT transformation)

(49) Obligation:

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

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

The TRS R consists of the following rules:

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

BEGIN(a(b(b(b(b(x)))))) → WAIT(Right1(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 \

| -I |

\ 1A /

+
/ 0A 0A 0A \

| 0A 0A 0A |

\ 1A 1A 1A /

· x₁

POL(BEGIN(x₁)) = 0A +
[ 0A, 0A, 1A ]

· x₁

POL(a(x₁)) =
/ 0A \

| 0A |

\ -I /

+
/ 0A -I 0A \

| 0A -I 0A |

\ 0A -I 0A /

· x₁

POL(b(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A -I 0A \

| 0A -I 0A |

\ 0A 0A 0A /

· x₁

POL(Right1(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A -I 0A \

| 0A -I 0A |

\ 0A -I 0A /

· x₁

POL(Right2(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A -I 1A \

| 0A -I 1A |

\ 1A -I 0A /

· x₁

POL(Right4(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A -I 1A \

| 1A -I 0A |

\ 0A -I 1A /

· x₁

POL(End(x₁)) =
/ -I \

| 1A |

\ -I /

+
/ 0A 0A -I \

| 1A 1A 0A |

\ 0A 0A -I /

· x₁

POL(Ab(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A 0A 0A \

| 0A 0A 0A |

\ 0A 0A 0A /

· x₁

POL(Aa(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A 0A 0A \

| 0A 0A 0A |

\ 0A 0A 0A /

· x₁

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

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

(51) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(b(b(b(b(x))))) → WAIT(Right2(x))
BEGIN(b(b(x))) → WAIT(Right4(x))

The TRS R consists of the following rules:

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

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

(52) UsableRulesProof (EQUIVALENT transformation)

(53) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(b(b(b(b(x))))) → WAIT(Right2(x))
BEGIN(b(b(x))) → WAIT(Right4(x))

The TRS R consists of the following rules:

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

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

(54) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

BEGIN(b(b(b(b(x))))) → WAIT(Right2(x))
BEGIN(b(b(x))) → WAIT(Right4(x))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x₁)) = 0
POL(Ab(x₁)) = 1 + x₁
POL(BEGIN(x₁)) = x₁
POL(End(x₁)) = x₁
POL(Left(x₁)) = x₁
POL(Right2(x₁)) = x₁
POL(Right4(x₁)) = x₁
POL(WAIT(x₁)) = x₁
POL(a(x₁)) = 0
POL(b(x₁)) = 1 + x₁

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

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

(55) Obligation:

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

WAIT(Left(x)) → BEGIN(x)

The TRS R consists of the following rules:

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

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

(56) DependencyGraphProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) UsableRulesProof (EQUIVALENT transformation)

(7) Obligation:

(8) QDPOrderProof (EQUIVALENT transformation)

(9) Obligation:

(10) QDPOrderProof (EQUIVALENT transformation)

(11) Obligation:

(12) PisEmptyProof (EQUIVALENT transformation)

(13) YES

(14) Obligation:

(15) UsableRulesProof (EQUIVALENT transformation)

(16) Obligation:

(17) QDPSizeChangeProof (EQUIVALENT transformation)

(18) YES

(19) Obligation:

(20) UsableRulesProof (EQUIVALENT transformation)

(21) Obligation:

(22) QDPSizeChangeProof (EQUIVALENT transformation)

(23) YES

(24) Obligation:

(25) UsableRulesProof (EQUIVALENT transformation)

(26) Obligation:

(27) QDPSizeChangeProof (EQUIVALENT transformation)

(28) YES

(29) Obligation:

(30) UsableRulesProof (EQUIVALENT transformation)

(31) Obligation:

(32) QDPSizeChangeProof (EQUIVALENT transformation)

(33) YES

(34) Obligation:

(35) UsableRulesProof (EQUIVALENT transformation)

(36) Obligation:

(37) QDPSizeChangeProof (EQUIVALENT transformation)

(38) YES

(39) Obligation:

(40) UsableRulesProof (EQUIVALENT transformation)

(41) Obligation:

(42) QDPOrderProof (EQUIVALENT transformation)

(43) Obligation:

(44) UsableRulesProof (EQUIVALENT transformation)

(45) Obligation:

(46) QDPOrderProof (EQUIVALENT transformation)

(47) Obligation:

(48) UsableRulesProof (EQUIVALENT transformation)

(49) Obligation:

(50) QDPOrderProof (EQUIVALENT transformation)

(51) Obligation:

(52) UsableRulesProof (EQUIVALENT transformation)

(53) Obligation:

(54) QDPOrderProof (EQUIVALENT transformation)

(55) Obligation:

(56) DependencyGraphProof (EQUIVALENT transformation)

(57) TRUE