(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
1(1(x)) → 4(3(x))
1(2(x)) → 2(1(x))
2(2(x)) → 1(1(1(x)))
3(3(x)) → 5(6(x))
3(4(x)) → 1(1(x))
4(4(x)) → 3(x)
5(5(x)) → 6(2(x))
5(6(x)) → 1(2(x))
6(6(x)) → 2(1(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(1(x)) → 3(4(x))
2(1(x)) → 1(2(x))
2(2(x)) → 1(1(1(x)))
3(3(x)) → 6(5(x))
4(3(x)) → 1(1(x))
4(4(x)) → 3(x)
5(5(x)) → 2(6(x))
6(5(x)) → 2(1(x))
6(6(x)) → 1(2(x))
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(1(x1)) = 85 + x1
POL(2(x1)) = 128 + x1
POL(3(x1)) = 113 + x1
POL(4(x1)) = 57 + x1
POL(5(x1)) = 118 + x1
POL(6(x1)) = 107 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
2(2(x)) → 1(1(1(x)))
3(3(x)) → 6(5(x))
4(4(x)) → 3(x)
5(5(x)) → 2(6(x))
6(5(x)) → 2(1(x))
6(6(x)) → 1(2(x))
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
1(1(x)) → 3(4(x))
2(1(x)) → 1(2(x))
4(3(x)) → 1(1(x))
Q is empty.
(5) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
11(1(x)) → 41(x)
21(1(x)) → 11(2(x))
21(1(x)) → 21(x)
41(3(x)) → 11(1(x))
41(3(x)) → 11(x)
The TRS R consists of the following rules:
1(1(x)) → 3(4(x))
2(1(x)) → 1(2(x))
4(3(x)) → 1(1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
(8) Complex Obligation (AND)
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
41(3(x)) → 11(1(x))
11(1(x)) → 41(x)
41(3(x)) → 11(x)
The TRS R consists of the following rules:
1(1(x)) → 3(4(x))
2(1(x)) → 1(2(x))
4(3(x)) → 1(1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) 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.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
41(3(x)) → 11(1(x))
11(1(x)) → 41(x)
41(3(x)) → 11(x)
The TRS R consists of the following rules:
1(1(x)) → 3(4(x))
4(3(x)) → 1(1(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.
41(3(x)) → 11(1(x))
41(3(x)) → 11(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( 11(x1) ) = max{0, 2x1 - 2} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
1(1(x)) → 3(4(x))
4(3(x)) → 1(1(x))
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
11(1(x)) → 41(x)
The TRS R consists of the following rules:
1(1(x)) → 3(4(x))
4(3(x)) → 1(1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(14) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(15) TRUE
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
21(1(x)) → 21(x)
The TRS R consists of the following rules:
1(1(x)) → 3(4(x))
2(1(x)) → 1(2(x))
4(3(x)) → 1(1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) 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.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
21(1(x)) → 21(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:
- 21(1(x)) → 21(x)
The graph contains the following edges 1 > 1
(20) YES