YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z126-split.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 52 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 MRRProof (⇔, 38 ms)
        ↳11 QDP
        ↳12 MRRProof (⇔, 15 ms)
        ↳13 QDP
        ↳14 MRRProof (⇔, 16 ms)
        ↳15 QDP
        ↳16 MRRProof (⇔, 13 ms)
        ↳17 QDP
        ↳18 QDPOrderProof (⇔, 58 ms)
        ↳19 QDP
        ↳20 PisEmptyProof (⇔, 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 QDPOrderProof (⇔, 30 ms)
        ↳31 QDP
        ↳32 MRRProof (⇔, 51 ms)
        ↳33 QDP
        ↳34 QDPOrderProof (⇔, 670 ms)
        ↳35 QDP
        ↳36 MRRProof (⇔, 0 ms)
        ↳37 QDP
        ↳38 MRRProof (⇔, 21 ms)
        ↳39 QDP
        ↳40 QDPOrderProof (⇔, 1138 ms)
        ↳41 QDP
        ↳42 MRRProof (⇔, 14 ms)
        ↳43 QDP
        ↳44 QDPOrderProof (⇔, 2737 ms)
        ↳45 QDP
        ↳46 PisEmptyProof (⇔, 0 ms)
        ↳47 YES

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

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

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

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

END(a(Right1(x))) → END(a(b(b(a(Left(x))))))
END(a(Right1(x))) → A(b(b(a(Left(x)))))
END(a(Right1(x))) → B(b(a(Left(x))))
END(a(Right1(x))) → B(a(Left(x)))
END(a(Right1(x))) → A(Left(x))
END(a(Right1(x))) → LEFT(x)
END(b(a(Right2(x)))) → END(a(b(b(a(Left(x))))))
END(b(a(Right2(x)))) → A(b(b(a(Left(x)))))
END(b(a(Right2(x)))) → B(b(a(Left(x))))
END(b(a(Right2(x)))) → B(a(Left(x)))
END(b(a(Right2(x)))) → A(Left(x))
END(b(a(Right2(x)))) → LEFT(x)
END(b(Right3(x))) → END(b(b(Left(x))))
END(b(Right3(x))) → B(b(Left(x)))
END(b(Right3(x))) → B(Left(x))
END(b(Right3(x))) → LEFT(x)
END(b(b(Right4(x)))) → END(b(b(Left(x))))
END(b(b(Right4(x)))) → B(b(Left(x)))
END(b(b(Right4(x)))) → B(Left(x))
END(b(b(Right4(x)))) → LEFT(x)
LEFT(Aa(x)) → A(Left(x))
LEFT(Aa(x)) → LEFT(x)
LEFT(Ab(x)) → B(Left(x))
LEFT(Ab(x)) → LEFT(x)
A(b(a(x))) → A(b(b(a(x))))
A(b(a(x))) → B(b(a(x)))

The TRS R consists of the following rules:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 19 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

A(b(a(x))) → A(b(b(a(x))))

The TRS R consists of the following rules:

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

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

(8) 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.

(9) Obligation:

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

A(b(a(x))) → A(b(b(a(x))))

The TRS R consists of the following rules:

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

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

(10) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = 2·x1   
POL(Aa(x1)) = x1   
POL(Ab(x1)) = x1   
POL(Begin(x1)) = 2 + 2·x1   
POL(Right1(x1)) = x1   
POL(Right2(x1)) = x1   
POL(Right3(x1)) = 2·x1   
POL(Right4(x1)) = x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = x1   

(11) Obligation:

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

A(b(a(x))) → A(b(b(a(x))))

The TRS R consists of the following rules:

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

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

(12) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

a(Right2(x)) → Right2(Aa(x))
a(Right4(x)) → Right4(Aa(x))

Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = x1   
POL(Aa(x1)) = x1   
POL(Ab(x1)) = x1   
POL(Right1(x1)) = x1   
POL(Right2(x1)) = 1 + 2·x1   
POL(Right3(x1)) = x1   
POL(Right4(x1)) = 3 + 2·x1   
POL(a(x1)) = 2·x1   
POL(b(x1)) = x1   

(13) Obligation:

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

A(b(a(x))) → A(b(b(a(x))))

The TRS R consists of the following rules:

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

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

(14) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

a(Right1(x)) → Right1(Aa(x))

Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = 2·x1   
POL(Aa(x1)) = x1   
POL(Ab(x1)) = x1   
POL(Right1(x1)) = 3 + 2·x1   
POL(Right2(x1)) = x1   
POL(Right3(x1)) = 2·x1   
POL(Right4(x1)) = 2·x1   
POL(a(x1)) = 2·x1   
POL(b(x1)) = x1   

(15) Obligation:

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

A(b(a(x))) → A(b(b(a(x))))

The TRS R consists of the following rules:

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

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

(16) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

a(Right3(x)) → Right3(Aa(x))

Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = 2·x1   
POL(Aa(x1)) = 1 + x1   
POL(Ab(x1)) = x1   
POL(Right1(x1)) = 2·x1   
POL(Right2(x1)) = 2·x1   
POL(Right3(x1)) = 2 + x1   
POL(Right4(x1)) = 2·x1   
POL(a(x1)) = 2·x1   
POL(b(x1)) = x1   

(17) Obligation:

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

A(b(a(x))) → A(b(b(a(x))))

The TRS R consists of the following rules:

a(b(a(x))) → a(b(b(a(x))))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
b(b(b(x))) → b(b(x))

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

(18) 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(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,-I,1A]
·x1

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

POL(a(x1)) =
/-I\
|1A|
\0A/
 +
/0A-I0A\
|1A-I1A|
\-I-I-I/
·x1

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

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

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

POL(Right3(x1)) =
/-I\
|0A|
\0A/
 +
/1A1A1A\
|0A-I0A|
\0A-I-I/
·x1

POL(Right4(x1)) =
/-I\
|-I|
\-I/
 +
/1A1A1A\
|0A1A1A|
\0A-I0A/
·x1

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

a(b(a(x))) → a(b(b(a(x))))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
b(b(b(x))) → b(b(x))

(19) Obligation:

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

a(b(a(x))) → a(b(b(a(x))))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
b(b(b(x))) → b(b(x))

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) YES

(22) Obligation:

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

LEFT(Ab(x)) → LEFT(x)
LEFT(Aa(x)) → LEFT(x)

The TRS R consists of the following rules:

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

LEFT(Ab(x)) → LEFT(x)
LEFT(Aa(x)) → LEFT(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:

  • LEFT(Ab(x)) → LEFT(x)
    The graph contains the following edges 1 > 1

  • LEFT(Aa(x)) → LEFT(x)
    The graph contains the following edges 1 > 1

(26) YES

(27) Obligation:

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

END(b(a(Right2(x)))) → END(a(b(b(a(Left(x))))))
END(a(Right1(x))) → END(a(b(b(a(Left(x))))))
END(b(Right3(x))) → END(b(b(Left(x))))
END(b(b(Right4(x)))) → END(b(b(Left(x))))

The TRS R consists of the following rules:

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

END(b(a(Right2(x)))) → END(a(b(b(a(Left(x))))))
END(a(Right1(x))) → END(a(b(b(a(Left(x))))))
END(b(Right3(x))) → END(b(b(Left(x))))
END(b(b(Right4(x)))) → END(b(b(Left(x))))

The TRS R consists of the following rules:

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

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

(30) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


END(b(a(Right2(x)))) → END(a(b(b(a(Left(x))))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x1)) = x1   
POL(Ab(x1)) = x1   
POL(Begin(x1)) = x1   
POL(END(x1)) = x1   
POL(Left(x1)) = 0   
POL(Right1(x1)) = 0   
POL(Right2(x1)) = 0   
POL(Right3(x1)) = 0   
POL(Right4(x1)) = 0   
POL(Wait(x1)) = x1   
POL(a(x1)) = 0   
POL(b(x1)) = 1   

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

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

(31) Obligation:

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

END(a(Right1(x))) → END(a(b(b(a(Left(x))))))
END(b(Right3(x))) → END(b(b(Left(x))))
END(b(b(Right4(x)))) → END(b(b(Left(x))))

The TRS R consists of the following rules:

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

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

(32) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

a(Begin(x)) → Right2(Wait(x))
a(Right2(x)) → Right2(Aa(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x1)) = 1 + x1   
POL(Ab(x1)) = x1   
POL(Begin(x1)) = 2·x1   
POL(END(x1)) = x1   
POL(Left(x1)) = 2·x1   
POL(Right1(x1)) = 2 + 2·x1   
POL(Right2(x1)) = x1   
POL(Right3(x1)) = 2·x1   
POL(Right4(x1)) = 2·x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = 2 + x1   
POL(b(x1)) = x1   

(33) Obligation:

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

END(a(Right1(x))) → END(a(b(b(a(Left(x))))))
END(b(Right3(x))) → END(b(b(Left(x))))
END(b(b(Right4(x)))) → END(b(b(Left(x))))

The TRS R consists of the following rules:

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

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

(34) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

POL(END(x1)) = 0A +
[-I,0A,-I]
·x1

POL(a(x1)) =
/-I\
|-I|
\-I/
 +
/0A-I0A\
|-I-I-I|
\0A0A1A/
·x1

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

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

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

POL(Right3(x1)) =
/-I\
|-I|
\1A/
 +
/0A-I0A\
|-I-I1A|
\1A0A-I/
·x1

POL(Right4(x1)) =
/1A\
|-I|
\0A/
 +
/0A-I0A\
|-I-I1A|
\1A-I-I/
·x1

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

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

POL(Wait(x1)) =
/0A\
|0A|
\1A/
 +
/-I0A-I\
|-I0A0A|
\-I1A0A/
·x1

POL(Begin(x1)) =
/0A\
|1A|
\-I/
 +
/-I0A0A\
|-I1A0A|
\-I0A-I/
·x1

POL(Right2(x1)) =
/0A\
|-I|
\1A/
 +
/0A0A-I\
|0A0A1A|
\0A0A0A/
·x1

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

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

(35) Obligation:

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

END(a(Right1(x))) → END(a(b(b(a(Left(x))))))
END(b(Right3(x))) → END(b(b(Left(x))))

The TRS R consists of the following rules:

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

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

(36) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

b(Begin(x)) → Right4(Wait(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x1)) = x1   
POL(Ab(x1)) = x1   
POL(Begin(x1)) = 2 + 2·x1   
POL(END(x1)) = 2·x1   
POL(Left(x1)) = 2·x1   
POL(Right1(x1)) = 2·x1   
POL(Right2(x1)) = 2·x1   
POL(Right3(x1)) = 2·x1   
POL(Right4(x1)) = x1   
POL(Wait(x1)) = 1 + x1   
POL(a(x1)) = x1   
POL(b(x1)) = x1   

(37) Obligation:

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

END(a(Right1(x))) → END(a(b(b(a(Left(x))))))
END(b(Right3(x))) → END(b(b(Left(x))))

The TRS R consists of the following rules:

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

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

(38) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

a(Right4(x)) → Right4(Aa(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x1)) = 2·x1   
POL(Ab(x1)) = x1   
POL(Begin(x1)) = x1   
POL(END(x1)) = 2·x1   
POL(Left(x1)) = x1   
POL(Right1(x1)) = 2·x1   
POL(Right2(x1)) = x1   
POL(Right3(x1)) = x1   
POL(Right4(x1)) = 2 + x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = 2·x1   
POL(b(x1)) = x1   

(39) Obligation:

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

END(a(Right1(x))) → END(a(b(b(a(Left(x))))))
END(b(Right3(x))) → END(b(b(Left(x))))

The TRS R consists of the following rules:

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

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

(40) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

POL(END(x1)) = 0A +
[0A,0A,-I]
·x1

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

POL(Right1(x1)) =
/1A\
|1A|
\0A/
 +
/1A-I0A\
|0A0A1A|
\-I1A-I/
·x1

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

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

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

POL(Aa(x1)) =
/0A\
|-I|
\0A/
 +
/0A0A1A\
|-I-I-I|
\-I-I0A/
·x1

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

POL(Wait(x1)) =
/1A\
|-I|
\0A/
 +
/-I1A0A\
|-I-I-I|
\-I0A-I/
·x1

POL(Begin(x1)) =
/1A\
|0A|
\0A/
 +
/-I1A0A\
|-I0A-I|
\-I-I-I/
·x1

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

POL(Right4(x1)) =
/0A\
|-I|
\-I/
 +
/-I-I0A\
|0A0A0A|
\0A0A0A/
·x1

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

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

(41) Obligation:

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

END(b(Right3(x))) → END(b(b(Left(x))))

The TRS R consists of the following rules:

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

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

(42) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

a(b(Begin(x))) → Right1(Wait(x))
a(Right1(x)) → Right1(Aa(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x1)) = 2·x1   
POL(Ab(x1)) = x1   
POL(Begin(x1)) = 2 + 2·x1   
POL(END(x1)) = 2·x1   
POL(Left(x1)) = 2·x1   
POL(Right1(x1)) = 1 + 2·x1   
POL(Right2(x1)) = x1   
POL(Right3(x1)) = 2·x1   
POL(Right4(x1)) = x1   
POL(Wait(x1)) = 1 + x1   
POL(a(x1)) = 2·x1   
POL(b(x1)) = x1   

(43) Obligation:

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

END(b(Right3(x))) → END(b(b(Left(x))))

The TRS R consists of the following rules:

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


END(b(Right3(x))) → END(b(b(Left(x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] to (N^3, +, *, >=, >) :

POL(END(x1)) = 0 +
[1,0,0]
·x1

POL(b(x1)) =
/0\
|0|
\1/
 +
/020\
|002|
\001/
·x1

POL(Right3(x1)) =
/0\
|2|
\2/
 +
/020\
|002|
\001/
·x1

POL(Left(x1)) =
/0\
|0|
\0/
 +
/020\
|002|
\001/
·x1

POL(Aa(x1)) =
/0\
|0|
\2/
 +
/000\
|000|
\023/
·x1

POL(a(x1)) =
/0\
|2|
\0/
 +
/000\
|222|
\111/
·x1

POL(Ab(x1)) =
/0\
|0|
\1/
 +
/000\
|002|
\001/
·x1

POL(Wait(x1)) =
/0\
|0|
\0/
 +
/000\
|000|
\000/
·x1

POL(Begin(x1)) =
/0\
|0|
\0/
 +
/000\
|000|
\000/
·x1

POL(Right1(x1)) =
/0\
|0|
\0/
 +
/000\
|000|
\201/
·x1

POL(Right2(x1)) =
/2\
|2|
\2/
 +
/000\
|000|
\001/
·x1

POL(Right4(x1)) =
/0\
|0|
\3/
 +
/000\
|000|
\200/
·x1

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

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

(45) Obligation:

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

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

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

(46) PisEmptyProof (EQUIVALENT transformation)

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

(47) YES