YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_06/16.srs-torpacyc2out-split.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 37 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPOrderProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 PisEmptyProof (⇔, 0 ms)
        ↳11 YES
    ↳12 QDP
        ↳13 UsableRulesProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 QDPSizeChangeProof (⇔, 0 ms)
        ↳16 YES
    ↳17 QDP
        ↳18 UsableRulesProof (⇔, 0 ms)
        ↳19 QDP
        ↳20 QDPSizeChangeProof (⇔, 0 ms)
        ↳21 YES
    ↳22 QDP
        ↳23 UsableRulesProof (⇔, 0 ms)
        ↳24 QDP
        ↳25 QDPSizeChangeProof (⇔, 0 ms)
        ↳26 YES
    ↳27 QDP
        ↳28 UsableRulesProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 QDPSizeChangeProof (⇔, 0 ms)
        ↳31 YES
    ↳32 QDP
        ↳33 UsableRulesProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 QDPOrderProof (⇔, 2675 ms)
        ↳36 QDP
        ↳37 UsableRulesProof (⇔, 0 ms)
        ↳38 QDP
        ↳39 QDPOrderProof (⇔, 2767 ms)
        ↳40 QDP
        ↳41 UsableRulesProof (⇔, 0 ms)
        ↳42 QDP
        ↳43 QDPOrderProof (⇔, 5697 ms)
        ↳44 QDP
        ↳45 UsableRulesProof (⇔, 0 ms)
        ↳46 QDP
        ↳47 QDPOrderProof (⇔, 1549 ms)
        ↳48 QDP
        ↳49 DependencyGraphProof (⇔, 0 ms)
        ↳50 TRUE

(0) Obligation:

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

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

The TRS R consists of the following rules:

Begin(b(a(x))) → Wait(Right1(x))
Begin(a(x)) → Wait(Right2(x))
Begin(b(a(x))) → Wait(Right3(x))
Begin(a(x)) → Wait(Right4(x))
Right1(a(End(x))) → Left(b(a(b(b(a(End(x)))))))
Right2(a(b(End(x)))) → Left(b(a(b(b(a(End(x)))))))
Right3(a(End(x))) → Left(a(b(b(a(b(End(x)))))))
Right4(a(b(End(x)))) → Left(a(b(b(a(b(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))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
a(b(a(x))) → b(a(b(b(a(x)))))
a(b(a(x))) → a(b(b(a(b(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 6 SCCs with 21 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

A(b(a(x))) → A(b(x))

The TRS R consists of the following rules:

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

R is empty.
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.


A(b(a(x))) → A(b(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(9) Obligation:

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) YES

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

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

(13) 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.

(14) 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.

(15) 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

(16) YES

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

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

(18) 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.

(19) 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.

(20) 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

(21) YES

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

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

(23) 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.

(24) 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.

(25) 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

(26) YES

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

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

(28) 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.

(29) 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.

(30) 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

(31) YES

(32) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(33) 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.

(34) Obligation:

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

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

The TRS R consists of the following rules:

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

(35) 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(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,0A,0A]
·x1

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

POL(BEGIN(x1)) = 0A +
[0A,1A,-I]
·x1

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

POL(a(x1)) =
/-I\
|0A|
\-I/
 +
/0A1A0A\
|0A1A0A|
\-I0A0A/
·x1

POL(Right1(x1)) =
/-I\
|-I|
\-I/
 +
/0A0A0A\
|0A1A0A|
\0A0A0A/
·x1

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

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

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

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

POL(Aa(x1)) =
/0A\
|0A|
\-I/
 +
/0A0A0A\
|-I1A-I|
\-I0A-I/
·x1

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

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

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

(36) Obligation:

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

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

The TRS R consists of the following rules:

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

(37) 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.

(38) Obligation:

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

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

The TRS R consists of the following rules:

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

(39) 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(Right2(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-I-I\
|0A0A0A|
\-I0A0A/
·x1

POL(BEGIN(x1)) = 0A +
[0A,0A,0A]
·x1

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

POL(a(x1)) =
/0A\
|0A|
\0A/
 +
/1A1A0A\
|0A0A0A|
\0A0A0A/
·x1

POL(Right1(x1)) =
/0A\
|-I|
\0A/
 +
/0A0A0A\
|0A0A-I|
\-I-I0A/
·x1

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

POL(Right3(x1)) =
/0A\
|-I|
\-I/
 +
/-I0A-I\
|0A0A-I|
\-I0A0A/
·x1

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

POL(Aa(x1)) =
/-I\
|0A|
\0A/
 +
/0A0A0A\
|0A1A0A|
\-I0A0A/
·x1

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

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

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

(40) Obligation:

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

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

The TRS R consists of the following rules:

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

(41) 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.

(42) Obligation:

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

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

The TRS R consists of the following rules:

Right3(a(End(x))) → Left(a(b(b(a(b(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(x))) → b(a(b(b(a(x)))))
a(b(a(x))) → a(b(b(a(b(x)))))
Right1(a(End(x))) → Left(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.

(43) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


BEGIN(b(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 +
[0A,0A,1A]
·x1

POL(Left(x1)) =
/0A\
|0A|
\0A/
 +
/-I-I-I\
|0A1A1A|
\0A0A1A/
·x1

POL(BEGIN(x1)) = 1A +
[1A,0A,0A]
·x1

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

POL(a(x1)) =
/0A\
|-I|
\-I/
 +
/-I-I1A\
|-I-I0A|
\-I0A0A/
·x1

POL(Right1(x1)) =
/0A\
|-I|
\-I/
 +
/-I0A0A\
|-I0A1A|
\-I-I1A/
·x1

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

POL(End(x1)) =
/0A\
|0A|
\-I/
 +
/0A-I0A\
|1A1A1A|
\0A0A0A/
·x1

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

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

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

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

(44) Obligation:

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

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

The TRS R consists of the following rules:

Right3(a(End(x))) → Left(a(b(b(a(b(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(x))) → b(a(b(b(a(x)))))
a(b(a(x))) → a(b(b(a(b(x)))))
Right1(a(End(x))) → Left(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.

(45) 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.

(46) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(47) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


BEGIN(b(a(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/
 +
/0A1A0A\
|-I0A1A|
\0A0A0A/
·x1

POL(BEGIN(x1)) = 0A +
[0A,1A,0A]
·x1

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

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

POL(Right1(x1)) =
/-I\
|-I|
\-I/
 +
/0A1A0A\
|0A-I1A|
\0A0A-I/
·x1

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

POL(Aa(x1)) =
/1A\
|0A|
\0A/
 +
/1A1A0A\
|1A1A0A|
\0A0A0A/
·x1

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

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

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

(48) 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:

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

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

(49) DependencyGraphProof (EQUIVALENT transformation)

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

(50) TRUE