(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
Begin(1(x)) → Wait(Right1(x))
Begin(4(x)) → Wait(Right2(x))
Begin(4(x)) → Wait(Right3(x))
Right1(2(End(x))) → Left(4(1(End(x))))
Right2(3(End(x))) → Left(4(2(End(x))))
Right3(2(End(x))) → Left(3(2(End(x))))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right1(3(x)) → A3(Right1(x))
Right2(3(x)) → A3(Right2(x))
Right3(3(x)) → A3(Right3(x))
A2(Left(x)) → Left(2(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A3(Left(x)) → Left(3(x))
Wait(Left(x)) → Begin(x)
2(1(x)) → 4(1(x))
3(4(x)) → 4(2(x))
2(4(x)) → 3(2(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:
1(Begin(x)) → Right1(Wait(x))
4(Begin(x)) → Right2(Wait(x))
4(Begin(x)) → Right3(Wait(x))
End(2(Right1(x))) → End(1(4(Left(x))))
End(3(Right2(x))) → End(2(4(Left(x))))
End(2(Right3(x))) → End(2(3(Left(x))))
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
2(Right3(x)) → Right3(A2(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
4(Right1(x)) → Right1(A4(x))
4(Right2(x)) → Right2(A4(x))
4(Right3(x)) → Right3(A4(x))
3(Right1(x)) → Right1(A3(x))
3(Right2(x)) → Right2(A3(x))
3(Right3(x)) → Right3(A3(x))
Left(A2(x)) → 2(Left(x))
Left(A1(x)) → 1(Left(x))
Left(A4(x)) → 4(Left(x))
Left(A3(x)) → 3(Left(x))
Left(Wait(x)) → Begin(x)
1(2(x)) → 1(4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(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(2(Right1(x))) → END(1(4(Left(x))))
END(2(Right1(x))) → 11(4(Left(x)))
END(2(Right1(x))) → 41(Left(x))
END(2(Right1(x))) → LEFT(x)
END(3(Right2(x))) → END(2(4(Left(x))))
END(3(Right2(x))) → 21(4(Left(x)))
END(3(Right2(x))) → 41(Left(x))
END(3(Right2(x))) → LEFT(x)
END(2(Right3(x))) → END(2(3(Left(x))))
END(2(Right3(x))) → 21(3(Left(x)))
END(2(Right3(x))) → 31(Left(x))
END(2(Right3(x))) → LEFT(x)
LEFT(A2(x)) → 21(Left(x))
LEFT(A2(x)) → LEFT(x)
LEFT(A1(x)) → 11(Left(x))
LEFT(A1(x)) → LEFT(x)
LEFT(A4(x)) → 41(Left(x))
LEFT(A4(x)) → LEFT(x)
LEFT(A3(x)) → 31(Left(x))
LEFT(A3(x)) → LEFT(x)
11(2(x)) → 11(4(x))
11(2(x)) → 41(x)
41(3(x)) → 21(4(x))
41(3(x)) → 41(x)
41(2(x)) → 21(3(x))
41(2(x)) → 31(x)
The TRS R consists of the following rules:
1(Begin(x)) → Right1(Wait(x))
4(Begin(x)) → Right2(Wait(x))
4(Begin(x)) → Right3(Wait(x))
End(2(Right1(x))) → End(1(4(Left(x))))
End(3(Right2(x))) → End(2(4(Left(x))))
End(2(Right3(x))) → End(2(3(Left(x))))
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
2(Right3(x)) → Right3(A2(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
4(Right1(x)) → Right1(A4(x))
4(Right2(x)) → Right2(A4(x))
4(Right3(x)) → Right3(A4(x))
3(Right1(x)) → Right1(A3(x))
3(Right2(x)) → Right2(A3(x))
3(Right3(x)) → Right3(A3(x))
Left(A2(x)) → 2(Left(x))
Left(A1(x)) → 1(Left(x))
Left(A4(x)) → 4(Left(x))
Left(A3(x)) → 3(Left(x))
Left(Wait(x)) → Begin(x)
1(2(x)) → 1(4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(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 4 SCCs with 17 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
41(3(x)) → 41(x)
The TRS R consists of the following rules:
1(Begin(x)) → Right1(Wait(x))
4(Begin(x)) → Right2(Wait(x))
4(Begin(x)) → Right3(Wait(x))
End(2(Right1(x))) → End(1(4(Left(x))))
End(3(Right2(x))) → End(2(4(Left(x))))
End(2(Right3(x))) → End(2(3(Left(x))))
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
2(Right3(x)) → Right3(A2(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
4(Right1(x)) → Right1(A4(x))
4(Right2(x)) → Right2(A4(x))
4(Right3(x)) → Right3(A4(x))
3(Right1(x)) → Right1(A3(x))
3(Right2(x)) → Right2(A3(x))
3(Right3(x)) → Right3(A3(x))
Left(A2(x)) → 2(Left(x))
Left(A1(x)) → 1(Left(x))
Left(A4(x)) → 4(Left(x))
Left(A3(x)) → 3(Left(x))
Left(Wait(x)) → Begin(x)
1(2(x)) → 1(4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(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:
41(3(x)) → 41(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) 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:
- 41(3(x)) → 41(x)
The graph contains the following edges 1 > 1
(11) YES
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
11(2(x)) → 11(4(x))
The TRS R consists of the following rules:
1(Begin(x)) → Right1(Wait(x))
4(Begin(x)) → Right2(Wait(x))
4(Begin(x)) → Right3(Wait(x))
End(2(Right1(x))) → End(1(4(Left(x))))
End(3(Right2(x))) → End(2(4(Left(x))))
End(2(Right3(x))) → End(2(3(Left(x))))
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
2(Right3(x)) → Right3(A2(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
4(Right1(x)) → Right1(A4(x))
4(Right2(x)) → Right2(A4(x))
4(Right3(x)) → Right3(A4(x))
3(Right1(x)) → Right1(A3(x))
3(Right2(x)) → Right2(A3(x))
3(Right3(x)) → Right3(A3(x))
Left(A2(x)) → 2(Left(x))
Left(A1(x)) → 1(Left(x))
Left(A4(x)) → 4(Left(x))
Left(A3(x)) → 3(Left(x))
Left(Wait(x)) → Begin(x)
1(2(x)) → 1(4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(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:
11(2(x)) → 11(4(x))
The TRS R consists of the following rules:
4(Begin(x)) → Right2(Wait(x))
4(Begin(x)) → Right3(Wait(x))
4(Right1(x)) → Right1(A4(x))
4(Right2(x)) → Right2(A4(x))
4(Right3(x)) → Right3(A4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(x))
3(Right1(x)) → Right1(A3(x))
3(Right2(x)) → Right2(A3(x))
3(Right3(x)) → Right3(A3(x))
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
2(Right3(x)) → Right3(A2(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) 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:
4(Begin(x)) → Right2(Wait(x))
4(Begin(x)) → Right3(Wait(x))
4(Right1(x)) → Right1(A4(x))
4(Right3(x)) → Right3(A4(x))
3(Right1(x)) → Right1(A3(x))
3(Right2(x)) → Right2(A3(x))
3(Right3(x)) → Right3(A3(x))
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
2(Right3(x)) → Right3(A2(x))
Used ordering: Polynomial interpretation [POLO]:
POL(11(x1)) = x1
POL(2(x1)) = 2 + 2·x1
POL(3(x1)) = 2 + 2·x1
POL(4(x1)) = 2 + 2·x1
POL(A2(x1)) = x1
POL(A3(x1)) = 1 + x1
POL(A4(x1)) = 2 + x1
POL(Begin(x1)) = 2 + 2·x1
POL(Right1(x1)) = 1 + x1
POL(Right2(x1)) = x1
POL(Right3(x1)) = 1 + x1
POL(Wait(x1)) = 2·x1
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
11(2(x)) → 11(4(x))
The TRS R consists of the following rules:
4(Right2(x)) → Right2(A4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
11(2(x)) → 11(4(x))
The TRS R consists of the following rules:
4(Right2(x)) → Right2(A4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(x))
The set Q consists of the following terms:
4(Right2(x0))
4(3(x0))
4(2(x0))
We have to consider all minimal (P,Q,R)-chains.
(19) 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 dependency pairs:
11(2(x)) → 11(4(x))
Used ordering: Polynomial interpretation [POLO]:
POL(11(x1)) = 2·x1
POL(2(x1)) = 2 + 2·x1
POL(3(x1)) = 1 + 2·x1
POL(4(x1)) = 2·x1
POL(A4(x1)) = x1
POL(Right2(x1)) = 2·x1
(20) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
4(Right2(x)) → Right2(A4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(x))
The set Q consists of the following terms:
4(Right2(x0))
4(3(x0))
4(2(x0))
We have to consider all minimal (P,Q,R)-chains.
(21) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(22) YES
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LEFT(A1(x)) → LEFT(x)
LEFT(A2(x)) → LEFT(x)
LEFT(A4(x)) → LEFT(x)
LEFT(A3(x)) → LEFT(x)
The TRS R consists of the following rules:
1(Begin(x)) → Right1(Wait(x))
4(Begin(x)) → Right2(Wait(x))
4(Begin(x)) → Right3(Wait(x))
End(2(Right1(x))) → End(1(4(Left(x))))
End(3(Right2(x))) → End(2(4(Left(x))))
End(2(Right3(x))) → End(2(3(Left(x))))
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
2(Right3(x)) → Right3(A2(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
4(Right1(x)) → Right1(A4(x))
4(Right2(x)) → Right2(A4(x))
4(Right3(x)) → Right3(A4(x))
3(Right1(x)) → Right1(A3(x))
3(Right2(x)) → Right2(A3(x))
3(Right3(x)) → Right3(A3(x))
Left(A2(x)) → 2(Left(x))
Left(A1(x)) → 1(Left(x))
Left(A4(x)) → 4(Left(x))
Left(A3(x)) → 3(Left(x))
Left(Wait(x)) → Begin(x)
1(2(x)) → 1(4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(24) 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.
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LEFT(A1(x)) → LEFT(x)
LEFT(A2(x)) → LEFT(x)
LEFT(A4(x)) → LEFT(x)
LEFT(A3(x)) → LEFT(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(26) 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(A1(x)) → LEFT(x)
The graph contains the following edges 1 > 1
- LEFT(A2(x)) → LEFT(x)
The graph contains the following edges 1 > 1
- LEFT(A4(x)) → LEFT(x)
The graph contains the following edges 1 > 1
- LEFT(A3(x)) → LEFT(x)
The graph contains the following edges 1 > 1
(27) YES
(28) Obligation:
Q DP problem:
The TRS P consists of the following rules:
END(3(Right2(x))) → END(2(4(Left(x))))
END(2(Right1(x))) → END(1(4(Left(x))))
END(2(Right3(x))) → END(2(3(Left(x))))
The TRS R consists of the following rules:
1(Begin(x)) → Right1(Wait(x))
4(Begin(x)) → Right2(Wait(x))
4(Begin(x)) → Right3(Wait(x))
End(2(Right1(x))) → End(1(4(Left(x))))
End(3(Right2(x))) → End(2(4(Left(x))))
End(2(Right3(x))) → End(2(3(Left(x))))
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
2(Right3(x)) → Right3(A2(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
4(Right1(x)) → Right1(A4(x))
4(Right2(x)) → Right2(A4(x))
4(Right3(x)) → Right3(A4(x))
3(Right1(x)) → Right1(A3(x))
3(Right2(x)) → Right2(A3(x))
3(Right3(x)) → Right3(A3(x))
Left(A2(x)) → 2(Left(x))
Left(A1(x)) → 1(Left(x))
Left(A4(x)) → 4(Left(x))
Left(A3(x)) → 3(Left(x))
Left(Wait(x)) → Begin(x)
1(2(x)) → 1(4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(29) 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.
(30) Obligation:
Q DP problem:
The TRS P consists of the following rules:
END(3(Right2(x))) → END(2(4(Left(x))))
END(2(Right1(x))) → END(1(4(Left(x))))
END(2(Right3(x))) → END(2(3(Left(x))))
The TRS R consists of the following rules:
Left(A2(x)) → 2(Left(x))
Left(A1(x)) → 1(Left(x))
Left(A4(x)) → 4(Left(x))
Left(A3(x)) → 3(Left(x))
Left(Wait(x)) → Begin(x)
3(Right1(x)) → Right1(A3(x))
3(Right2(x)) → Right2(A3(x))
3(Right3(x)) → Right3(A3(x))
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
2(Right3(x)) → Right3(A2(x))
4(Begin(x)) → Right2(Wait(x))
4(Begin(x)) → Right3(Wait(x))
4(Right1(x)) → Right1(A4(x))
4(Right2(x)) → Right2(A4(x))
4(Right3(x)) → Right3(A4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(x))
1(Begin(x)) → Right1(Wait(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
1(2(x)) → 1(4(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(31) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
END(3(Right2(x))) → END(2(4(Left(x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(1(x1)) = 0
POL(2(x1)) = 0
POL(3(x1)) = 1
POL(4(x1)) = 0
POL(A1(x1)) = x1
POL(A2(x1)) = x1
POL(A3(x1)) = x1
POL(A4(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(Wait(x1)) = x1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
2(Right3(x)) → Right3(A2(x))
1(Begin(x)) → Right1(Wait(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
1(2(x)) → 1(4(x))
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
END(2(Right1(x))) → END(1(4(Left(x))))
END(2(Right3(x))) → END(2(3(Left(x))))
The TRS R consists of the following rules:
Left(A2(x)) → 2(Left(x))
Left(A1(x)) → 1(Left(x))
Left(A4(x)) → 4(Left(x))
Left(A3(x)) → 3(Left(x))
Left(Wait(x)) → Begin(x)
3(Right1(x)) → Right1(A3(x))
3(Right2(x)) → Right2(A3(x))
3(Right3(x)) → Right3(A3(x))
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
2(Right3(x)) → Right3(A2(x))
4(Begin(x)) → Right2(Wait(x))
4(Begin(x)) → Right3(Wait(x))
4(Right1(x)) → Right1(A4(x))
4(Right2(x)) → Right2(A4(x))
4(Right3(x)) → Right3(A4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(x))
1(Begin(x)) → Right1(Wait(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
1(2(x)) → 1(4(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(33) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
END(2(Right1(x))) → END(1(4(Left(x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(1(x1)) = 0
POL(2(x1)) = 1
POL(3(x1)) = 0
POL(4(x1)) = 0
POL(A1(x1)) = x1
POL(A2(x1)) = x1
POL(A3(x1)) = x1
POL(A4(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(Wait(x1)) = x1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
1(Begin(x)) → Right1(Wait(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
1(2(x)) → 1(4(x))
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
2(Right3(x)) → Right3(A2(x))
(34) Obligation:
Q DP problem:
The TRS P consists of the following rules:
END(2(Right3(x))) → END(2(3(Left(x))))
The TRS R consists of the following rules:
Left(A2(x)) → 2(Left(x))
Left(A1(x)) → 1(Left(x))
Left(A4(x)) → 4(Left(x))
Left(A3(x)) → 3(Left(x))
Left(Wait(x)) → Begin(x)
3(Right1(x)) → Right1(A3(x))
3(Right2(x)) → Right2(A3(x))
3(Right3(x)) → Right3(A3(x))
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
2(Right3(x)) → Right3(A2(x))
4(Begin(x)) → Right2(Wait(x))
4(Begin(x)) → Right3(Wait(x))
4(Right1(x)) → Right1(A4(x))
4(Right2(x)) → Right2(A4(x))
4(Right3(x)) → Right3(A4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(x))
1(Begin(x)) → Right1(Wait(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
1(2(x)) → 1(4(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(35) 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:
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
Used ordering: Polynomial interpretation [POLO]:
POL(1(x1)) = 2 + x1
POL(2(x1)) = x1
POL(3(x1)) = x1
POL(4(x1)) = x1
POL(A1(x1)) = 1 + x1
POL(A2(x1)) = x1
POL(A3(x1)) = x1
POL(A4(x1)) = x1
POL(Begin(x1)) = 2·x1
POL(END(x1)) = 2·x1
POL(Left(x1)) = 2·x1
POL(Right1(x1)) = 2 + x1
POL(Right2(x1)) = x1
POL(Right3(x1)) = 2·x1
POL(Wait(x1)) = x1
(36) Obligation:
Q DP problem:
The TRS P consists of the following rules:
END(2(Right3(x))) → END(2(3(Left(x))))
The TRS R consists of the following rules:
Left(A2(x)) → 2(Left(x))
Left(A1(x)) → 1(Left(x))
Left(A4(x)) → 4(Left(x))
Left(A3(x)) → 3(Left(x))
Left(Wait(x)) → Begin(x)
3(Right1(x)) → Right1(A3(x))
3(Right2(x)) → Right2(A3(x))
3(Right3(x)) → Right3(A3(x))
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
2(Right3(x)) → Right3(A2(x))
4(Begin(x)) → Right2(Wait(x))
4(Begin(x)) → Right3(Wait(x))
4(Right1(x)) → Right1(A4(x))
4(Right2(x)) → Right2(A4(x))
4(Right3(x)) → Right3(A4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(x))
1(Begin(x)) → Right1(Wait(x))
1(Right3(x)) → Right3(A1(x))
1(2(x)) → 1(4(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(37) 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:
1(Begin(x)) → Right1(Wait(x))
Used ordering: Polynomial interpretation [POLO]:
POL(1(x1)) = 3·x1
POL(2(x1)) = 2·x1
POL(3(x1)) = 2·x1
POL(4(x1)) = 2·x1
POL(A1(x1)) = 3·x1
POL(A2(x1)) = 2·x1
POL(A3(x1)) = 2·x1
POL(A4(x1)) = 2·x1
POL(Begin(x1)) = 2 + x1
POL(END(x1)) = 2·x1
POL(Left(x1)) = x1
POL(Right1(x1)) = 2·x1
POL(Right2(x1)) = 2·x1
POL(Right3(x1)) = 2·x1
POL(Wait(x1)) = 2 + x1
(38) Obligation:
Q DP problem:
The TRS P consists of the following rules:
END(2(Right3(x))) → END(2(3(Left(x))))
The TRS R consists of the following rules:
Left(A2(x)) → 2(Left(x))
Left(A1(x)) → 1(Left(x))
Left(A4(x)) → 4(Left(x))
Left(A3(x)) → 3(Left(x))
Left(Wait(x)) → Begin(x)
3(Right1(x)) → Right1(A3(x))
3(Right2(x)) → Right2(A3(x))
3(Right3(x)) → Right3(A3(x))
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
2(Right3(x)) → Right3(A2(x))
4(Begin(x)) → Right2(Wait(x))
4(Begin(x)) → Right3(Wait(x))
4(Right1(x)) → Right1(A4(x))
4(Right2(x)) → Right2(A4(x))
4(Right3(x)) → Right3(A4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(x))
1(Right3(x)) → Right3(A1(x))
1(2(x)) → 1(4(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(39) 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:
4(Begin(x)) → Right2(Wait(x))
Used ordering: Polynomial interpretation [POLO]:
POL(1(x1)) = x1
POL(2(x1)) = x1
POL(3(x1)) = x1
POL(4(x1)) = x1
POL(A1(x1)) = x1
POL(A2(x1)) = x1
POL(A3(x1)) = x1
POL(A4(x1)) = x1
POL(Begin(x1)) = 2 + 2·x1
POL(END(x1)) = 2·x1
POL(Left(x1)) = 2·x1
POL(Right1(x1)) = x1
POL(Right2(x1)) = x1
POL(Right3(x1)) = 2·x1
POL(Wait(x1)) = 1 + x1
(40) Obligation:
Q DP problem:
The TRS P consists of the following rules:
END(2(Right3(x))) → END(2(3(Left(x))))
The TRS R consists of the following rules:
Left(A2(x)) → 2(Left(x))
Left(A1(x)) → 1(Left(x))
Left(A4(x)) → 4(Left(x))
Left(A3(x)) → 3(Left(x))
Left(Wait(x)) → Begin(x)
3(Right1(x)) → Right1(A3(x))
3(Right2(x)) → Right2(A3(x))
3(Right3(x)) → Right3(A3(x))
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
2(Right3(x)) → Right3(A2(x))
4(Begin(x)) → Right3(Wait(x))
4(Right1(x)) → Right1(A4(x))
4(Right2(x)) → Right2(A4(x))
4(Right3(x)) → Right3(A4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(x))
1(Right3(x)) → Right3(A1(x))
1(2(x)) → 1(4(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(41) 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:
3(Right1(x)) → Right1(A3(x))
3(Right2(x)) → Right2(A3(x))
2(Right1(x)) → Right1(A2(x))
2(Right2(x)) → Right2(A2(x))
4(Right1(x)) → Right1(A4(x))
4(Right2(x)) → Right2(A4(x))
Used ordering: Polynomial interpretation [POLO]:
POL(1(x1)) = 2·x1
POL(2(x1)) = 2·x1
POL(3(x1)) = 2·x1
POL(4(x1)) = 2·x1
POL(A1(x1)) = 2·x1
POL(A2(x1)) = 2·x1
POL(A3(x1)) = 2·x1
POL(A4(x1)) = 2·x1
POL(Begin(x1)) = 2·x1
POL(END(x1)) = 2·x1
POL(Left(x1)) = x1
POL(Right1(x1)) = 3 + 2·x1
POL(Right2(x1)) = 2 + 2·x1
POL(Right3(x1)) = 2·x1
POL(Wait(x1)) = 2·x1
(42) Obligation:
Q DP problem:
The TRS P consists of the following rules:
END(2(Right3(x))) → END(2(3(Left(x))))
The TRS R consists of the following rules:
Left(A2(x)) → 2(Left(x))
Left(A1(x)) → 1(Left(x))
Left(A4(x)) → 4(Left(x))
Left(A3(x)) → 3(Left(x))
Left(Wait(x)) → Begin(x)
3(Right3(x)) → Right3(A3(x))
2(Right3(x)) → Right3(A2(x))
4(Begin(x)) → Right3(Wait(x))
4(Right3(x)) → Right3(A4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(x))
1(Right3(x)) → Right3(A1(x))
1(2(x)) → 1(4(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(43) 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:
1(Right3(x)) → Right3(A1(x))
Used ordering: Polynomial interpretation [POLO]:
POL(1(x1)) = 2·x1
POL(2(x1)) = 1 + 2·x1
POL(3(x1)) = 1 + 2·x1
POL(4(x1)) = 1 + 2·x1
POL(A1(x1)) = 2·x1
POL(A2(x1)) = 1 + 2·x1
POL(A3(x1)) = 1 + 2·x1
POL(A4(x1)) = 1 + 2·x1
POL(Begin(x1)) = x1
POL(END(x1)) = 2·x1
POL(Left(x1)) = x1
POL(Right3(x1)) = 1 + 2·x1
POL(Wait(x1)) = x1
(44) Obligation:
Q DP problem:
The TRS P consists of the following rules:
END(2(Right3(x))) → END(2(3(Left(x))))
The TRS R consists of the following rules:
Left(A2(x)) → 2(Left(x))
Left(A1(x)) → 1(Left(x))
Left(A4(x)) → 4(Left(x))
Left(A3(x)) → 3(Left(x))
Left(Wait(x)) → Begin(x)
3(Right3(x)) → Right3(A3(x))
2(Right3(x)) → Right3(A2(x))
4(Begin(x)) → Right3(Wait(x))
4(Right3(x)) → Right3(A4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(x))
1(2(x)) → 1(4(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(45) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
END(2(Right3(x))) → END(2(3(Left(x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( END(x1) ) = max{0, 2x1 - 2} |
| POL( 1(x1) ) = max{0, -2} |
| POL( Right3(x1) ) = 2x1 + 2 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
Left(A2(x)) → 2(Left(x))
Left(A1(x)) → 1(Left(x))
Left(A4(x)) → 4(Left(x))
Left(A3(x)) → 3(Left(x))
Left(Wait(x)) → Begin(x)
3(Right3(x)) → Right3(A3(x))
2(Right3(x)) → Right3(A2(x))
1(2(x)) → 1(4(x))
4(Begin(x)) → Right3(Wait(x))
4(Right3(x)) → Right3(A4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(x))
(46) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
Left(A2(x)) → 2(Left(x))
Left(A1(x)) → 1(Left(x))
Left(A4(x)) → 4(Left(x))
Left(A3(x)) → 3(Left(x))
Left(Wait(x)) → Begin(x)
3(Right3(x)) → Right3(A3(x))
2(Right3(x)) → Right3(A2(x))
4(Begin(x)) → Right3(Wait(x))
4(Right3(x)) → Right3(A4(x))
4(3(x)) → 2(4(x))
4(2(x)) → 2(3(x))
1(2(x)) → 1(4(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(47) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(48) YES