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

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

0 QTRS

↳1 DependencyPairsProof (⇔, 0 ms)

↳2 QDP

↳3 DependencyGraphProof (⇔, 0 ms)

↳4 AND

    ↳5 QDP

        ↳6 UsableRulesProof (⇔, 0 ms)

        ↳7 QDP

        ↳8 QDPOrderProof (⇔, 108 ms)

        ↳9 QDP

        ↳10 DependencyGraphProof (⇔, 0 ms)

        ↳11 QDP

        ↳12 QDPOrderProof (⇔, 40 ms)

        ↳13 QDP

        ↳14 PisEmptyProof (⇔, 0 ms)

        ↳15 YES

    ↳16 QDP

        ↳17 UsableRulesProof (⇔, 0 ms)

        ↳18 QDP

        ↳19 QDPSizeChangeProof (⇔, 0 ms)

        ↳20 YES

    ↳21 QDP

        ↳22 UsableRulesProof (⇔, 0 ms)

        ↳23 QDP

        ↳24 QDPSizeChangeProof (⇔, 0 ms)

        ↳25 YES

    ↳26 QDP

        ↳27 UsableRulesProof (⇔, 0 ms)

        ↳28 QDP

        ↳29 QDPSizeChangeProof (⇔, 0 ms)

        ↳30 YES

    ↳31 QDP

        ↳32 UsableRulesProof (⇔, 0 ms)

        ↳33 QDP

        ↳34 QDPSizeChangeProof (⇔, 0 ms)

        ↳35 YES

    ↳36 QDP

        ↳37 UsableRulesProof (⇔, 0 ms)

        ↳38 QDP

        ↳39 QDPSizeChangeProof (⇔, 0 ms)

        ↳40 YES

    ↳41 QDP

        ↳42 UsableRulesProof (⇔, 0 ms)

        ↳43 QDP

        ↳44 QDPOrderProof (⇔, 2335 ms)

        ↳45 QDP

        ↳46 UsableRulesProof (⇔, 0 ms)

        ↳47 QDP

        ↳48 QDPOrderProof (⇔, 2641 ms)

        ↳49 QDP

        ↳50 UsableRulesProof (⇔, 0 ms)

        ↳51 QDP

        ↳52 QDPOrderProof (⇔, 1811 ms)

        ↳53 QDP

        ↳54 UsableRulesProof (⇔, 0 ms)

        ↳55 QDP

        ↳56 QDPOrderProof (⇔, 233 ms)

        ↳57 QDP

        ↳58 UsableRulesProof (⇔, 0 ms)

        ↳59 QDP

        ↳60 QDPOrderProof (⇔, 613 ms)

        ↳61 QDP

        ↳62 DependencyGraphProof (⇔, 0 ms)

        ↳63 TRUE

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

The TRS R consists of the following rules:

Begin(b(b(x))) → Wait(Right1(x))
Begin(b(x)) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Begin(a(a(x))) → Wait(Right4(x))
Begin(a(x)) → Wait(Right5(x))
Right1(b(End(x))) → Left(a(End(x)))
Right2(b(b(End(x)))) → Left(a(End(x)))
Right3(a(End(x))) → Left(a(b(a(End(x)))))
Right4(a(End(x))) → Left(b(a(a(End(x)))))
Right5(a(a(End(x)))) → Left(b(a(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(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
a(a(a(x))) → b(a(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 24 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

Begin(b(b(x))) → Wait(Right1(x))
Begin(b(x)) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Begin(a(a(x))) → Wait(Right4(x))
Begin(a(x)) → Wait(Right5(x))
Right1(b(End(x))) → Left(a(End(x)))
Right2(b(b(End(x)))) → Left(a(End(x)))
Right3(a(End(x))) → Left(a(b(a(End(x)))))
Right4(a(End(x))) → Left(b(a(a(End(x)))))
Right5(a(a(End(x)))) → Left(b(a(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(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))
a(a(a(x))) → b(a(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(a(x)) → A(b(a(x)))
A(a(x)) → B(a(x))
B(b(b(x))) → A(x)
A(a(a(x))) → B(a(a(x)))

The TRS R consists of the following rules:

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

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

POL(A(x₁)) = 0A +
[ 0A, 0A, 0A ]

· x₁

POL(a(x₁)) =
/ 0A \

| -I |

\ 0A /

+
/ 1A 1A 1A \

| 0A 0A 0A |

\ 1A -I -I /

· x₁

POL(b(x₁)) =
/ 0A \

| 0A |

\ -I /

+
/ -I -I 0A \

| 0A 1A 0A |

\ 0A 1A 0A /

· x₁

POL(B(x₁)) = 0A +
[ 0A, 0A, -I ]

· x₁

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

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

(9) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(10) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(11) Obligation:

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

A(a(x)) → A(b(a(x)))

The TRS R consists of the following rules:

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

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

(12) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

A(a(x)) → A(b(a(x)))

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

POL(A(x₁)) = -I +
[ 0A, 0A, -I ]

· x₁

POL(a(x₁)) =
/ 0A \

| 1A |

\ 0A /

+
/ -I 0A -I \

| 1A 1A 0A |

\ 0A 0A -I /

· x₁

POL(b(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A -I 0A \

| 0A -I 0A |

\ 0A 0A 1A /

· x₁

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

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

(13) Obligation:

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

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

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

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) YES

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

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

(17) UsableRulesProof (EQUIVALENT transformation)

(18) 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.

(19) 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

(20) YES

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

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

(22) UsableRulesProof (EQUIVALENT transformation)

(23) 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.

(24) 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

(25) YES

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

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

(27) UsableRulesProof (EQUIVALENT transformation)

(28) 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.

(29) 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

(30) YES

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

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

(32) UsableRulesProof (EQUIVALENT transformation)

(33) 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.

(34) 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

(35) YES

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

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

(37) UsableRulesProof (EQUIVALENT transformation)

(38) 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.

(39) 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

(40) YES

(41) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(42) UsableRulesProof (EQUIVALENT transformation)

(43) Obligation:

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

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

The TRS R consists of the following rules:

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

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

The following pairs can be oriented strictly and are deleted.

BEGIN(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 \

| 0A |

\ 0A /

+
/ 0A 1A 0A \

| 0A 0A 0A |

\ 1A 0A 0A /

· x₁

POL(BEGIN(x₁)) = 0A +
[ -I, 1A, 0A ]

· x₁

POL(b(x₁)) =
/ -I \

| -I |

\ 0A /

+
/ 0A 0A -I \

| 0A 0A -I |

\ -I 0A 0A /

· x₁

POL(Right1(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A -I -I \

| 0A -I -I |

\ 0A -I -I /

· x₁

POL(Right2(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A 0A 0A \

| 0A 0A 0A |

\ 0A 0A 0A /

· x₁

POL(a(x₁)) =
/ -I \

| -I |

\ 0A /

+
/ 0A -I -I \

| 0A -I -I |

\ 0A 0A 0A /

· x₁

POL(Right3(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A 0A 0A \

| 0A 0A 0A |

\ 0A 0A 0A /

· x₁

POL(Right4(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A 0A 0A \

| 0A -I 0A |

\ 0A 0A 0A /

· x₁

POL(Right5(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ -I 0A 0A \

| -I 0A 0A |

\ -I 0A 0A /

· x₁

POL(End(x₁)) =
/ -I \

| 0A |

\ 0A /

+
/ 0A 0A -I \

| 1A 1A 0A |

\ 0A -I 0A /

· x₁

POL(Ab(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A 0A 0A \

| 0A -I 0A |

\ 0A 0A 0A /

· x₁

POL(Aa(x₁)) =
/ -I \

| -I |

\ -I /

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

(45) Obligation:

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

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

The TRS R consists of the following rules:

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

(47) Obligation:

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

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

The TRS R consists of the following rules:

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

(48) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

BEGIN(a(a(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)) = 0A +
[ 0A, -I, 0A ]

· x₁

POL(Left(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ -I -I -I \

| -I 0A 0A |

\ 0A 0A -I /

· x₁

POL(BEGIN(x₁)) = 0A +
[ 0A, 0A, -I ]

· x₁

POL(b(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A 0A 0A \

| 0A 0A -I |

\ 0A -I 1A /

· x₁

POL(Right2(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A -I 0A \

| -I -I 0A |

\ 0A -I -I /

· x₁

POL(a(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 1A 0A 0A \

| 0A 0A -I |

\ 0A 0A -I /

· x₁

POL(Right3(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A 0A 0A \

| 0A -I 1A |

\ 1A 0A 0A /

· x₁

POL(Right4(x₁)) =
/ 0A \

| -I |

\ 0A /

+
/ 0A 0A 0A \

| 0A -I 1A |

\ 1A 0A -I /

· x₁

POL(Right5(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A -I 0A \

| 0A -I 1A |

\ 1A 0A 0A /

· x₁

POL(End(x₁)) =
/ -I \

| 0A |

\ 0A /

+
/ 0A -I -I \

| 1A -I 0A |

\ -I -I 0A /

· x₁

POL(Ab(x₁)) =
/ 0A \

| -I |

\ 0A /

+
/ -I 0A -I \

| 0A 1A 0A |

\ 0A 0A 0A /

· x₁

POL(Aa(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A -I 0A \

| -I -I 0A |

\ -I 0A 1A /

· x₁

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

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

(49) Obligation:

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

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

The TRS R consists of the following rules:

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

(50) UsableRulesProof (EQUIVALENT transformation)

(51) Obligation:

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

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

The TRS R consists of the following rules:

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

(52) 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(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)) =
/ 0A \

| 1A |

\ -I /

+
/ 0A 0A 0A \

| 0A 0A 1A |

\ 1A 1A 0A /

· x₁

POL(BEGIN(x₁)) = 0A +
[ 0A, 1A, 0A ]

· x₁

POL(b(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A 0A 0A \

| -I 0A 0A |

\ 0A 0A 0A /

· x₁

POL(Right2(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ -I 0A 0A \

| -I 0A 0A |

\ -I 0A 0A /

· x₁

POL(a(x₁)) =
/ 0A \

| -I |

\ 0A /

+
/ -I 0A 0A \

| -I -I 0A |

\ -I 0A 0A /

· x₁

POL(Right3(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ -I 0A 0A \

| -I 0A 1A |

\ -I 0A 1A /

· x₁

POL(Right5(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ -I 0A 0A \

| -I 0A 1A |

\ -I 0A 1A /

· x₁

POL(End(x₁)) =
/ 1A \

| -I |

\ -I /

+
/ 1A -I 0A \

| 0A -I -I |

\ 0A -I -I /

· x₁

POL(Ab(x₁)) =
/ 0A \

| 0A |

\ 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:

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

(53) Obligation:

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

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

The TRS R consists of the following rules:

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

(55) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(56) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

BEGIN(a(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 +
[ -I, 0A, -I ]

· x₁

POL(Left(x₁)) =
/ 1A \

| 1A |

\ 0A /

+
/ 0A -I 0A \

| -I 0A 0A |

\ -I -I 0A /

· x₁

POL(BEGIN(x₁)) = 1A +
[ -I, 0A, 0A ]

· x₁

POL(a(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ -I 0A 0A \

| -I 1A 0A |

\ -I 0A 0A /

· x₁

POL(Right3(x₁)) =
/ 0A \

| -I |

\ 0A /

+
/ 0A -I -I \

| -I 0A -I |

\ -I -I 0A /

· x₁

POL(Right5(x₁)) =
/ 0A \

| -I |

\ 1A /

+
/ 1A -I 0A \

| -I 1A 0A |

\ -I -I 1A /

· x₁

POL(End(x₁)) =
/ 0A \

| 1A |

\ 0A /

+
/ 0A 0A 0A \

| 0A 0A 0A |

\ 0A -I 0A /

· x₁

POL(b(x₁)) =
/ 1A \

| 0A |

\ 0A /

+
/ 1A 0A 0A \

| 0A -I 0A |

\ 0A -I 0A /

· x₁

POL(Ab(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 1A 0A 0A \

| 0A -I 0A |

\ 0A -I 0A /

· x₁

POL(Aa(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ -I 0A 0A \

| -I 1A 0A |

\ -I 0A 0A /

· x₁

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

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

(57) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(x)) → WAIT(Right5(x))

The TRS R consists of the following rules:

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

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

(58) UsableRulesProof (EQUIVALENT transformation)

(59) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(x)) → WAIT(Right5(x))

The TRS R consists of the following rules:

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

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

(60) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

BEGIN(a(x)) → WAIT(Right5(x))

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic integers [ARCTIC,STERNAGEL_THIEMANN_RTA14]:

POL(WAIT(x₁)) = -I +
[ 0A, -I, -1A ]

· x₁

POL(Left(x₁)) =
/ 0A \

| -I |

\ -1A /

+
/ -1A -1A -1A \

| -I 0A -I |

\ -1A -1A -1A /

· x₁

POL(BEGIN(x₁)) = 0A +
[ -1A, -I, -1A ]

· x₁

POL(a(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 2A -1A 1A \

| -1A -I -1A |

\ -1A -1A -1A /

· x₁

POL(Right5(x₁)) =
/ -I \

| -I |

\ -1A /

+
/ 0A -I -I \

| -1A 1A -1A |

\ 1A -I -1A /

· x₁

POL(End(x₁)) =
/ -I \

| -1A |

\ -I /

+
/ 0A -1A -I \

| -I -1A -1A |

\ -1A -1A -1A /

· x₁

POL(b(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A 2A 0A \

| -1A 1A -1A |

\ -I -1A -1A /

· x₁

POL(Ab(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A 1A -I \

| 0A 1A -I |

\ 0A 1A -1A /

· x₁

POL(Aa(x₁)) =
/ -I \

| -I |

\ -1A /

+
/ 2A -I -I \

| 0A -I -I |

\ 2A -1A -1A /

· x₁

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

Right5(a(a(End(x)))) → Left(b(a(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))
a(a(a(x))) → b(a(a(x)))
b(b(b(x))) → a(x)
a(a(x)) → a(b(a(x)))

(61) 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:

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

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

(62) 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) DependencyGraphProof (EQUIVALENT transformation)

(11) Obligation:

(12) QDPOrderProof (EQUIVALENT transformation)

(13) Obligation:

(14) PisEmptyProof (EQUIVALENT transformation)

(15) YES

(16) Obligation:

(17) UsableRulesProof (EQUIVALENT transformation)

(18) Obligation:

(19) QDPSizeChangeProof (EQUIVALENT transformation)

(20) YES

(21) Obligation:

(22) UsableRulesProof (EQUIVALENT transformation)

(23) Obligation:

(24) QDPSizeChangeProof (EQUIVALENT transformation)

(25) YES

(26) Obligation:

(27) UsableRulesProof (EQUIVALENT transformation)

(28) Obligation:

(29) QDPSizeChangeProof (EQUIVALENT transformation)

(30) YES

(31) Obligation:

(32) UsableRulesProof (EQUIVALENT transformation)

(33) Obligation:

(34) QDPSizeChangeProof (EQUIVALENT transformation)

(35) YES

(36) Obligation:

(37) UsableRulesProof (EQUIVALENT transformation)

(38) Obligation:

(39) QDPSizeChangeProof (EQUIVALENT transformation)

(40) YES

(41) Obligation:

(42) UsableRulesProof (EQUIVALENT transformation)

(43) Obligation:

(44) QDPOrderProof (EQUIVALENT transformation)

(45) Obligation:

(46) UsableRulesProof (EQUIVALENT transformation)

(47) Obligation:

(48) QDPOrderProof (EQUIVALENT transformation)

(49) Obligation:

(50) UsableRulesProof (EQUIVALENT transformation)

(51) Obligation:

(52) QDPOrderProof (EQUIVALENT transformation)

(53) Obligation:

(54) UsableRulesProof (EQUIVALENT transformation)

(55) Obligation:

(56) QDPOrderProof (EQUIVALENT transformation)

(57) Obligation:

(58) UsableRulesProof (EQUIVALENT transformation)

(59) Obligation:

(60) QDPOrderProof (EQUIVALENT transformation)

(61) Obligation:

(62) DependencyGraphProof (EQUIVALENT transformation)

(63) TRUE