(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
0(0(0(0(x)))) → 0(1(1(1(x))))
1(0(1(1(x)))) → 0(0(0(1(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:
01(0(0(0(x)))) → 01(1(1(1(x))))
01(0(0(0(x)))) → 11(1(1(x)))
01(0(0(0(x)))) → 11(1(x))
01(0(0(0(x)))) → 11(x)
11(0(1(1(x)))) → 01(0(0(1(x))))
11(0(1(1(x)))) → 01(0(1(x)))
11(0(1(1(x)))) → 01(1(x))
The TRS R consists of the following rules:
0(0(0(0(x)))) → 0(1(1(1(x))))
1(0(1(1(x)))) → 0(0(0(1(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
01(0(0(0(x)))) → 11(1(x))
01(0(0(0(x)))) → 11(x)
11(0(1(1(x)))) → 01(0(0(1(x))))
11(0(1(1(x)))) → 01(0(1(x)))
11(0(1(1(x)))) → 01(1(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0(x1)) = 1 + x1
POL(01(x1)) = x1
POL(1(x1)) = 1 + x1
POL(11(x1)) = 1 + x1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
1(0(1(1(x)))) → 0(0(0(1(x))))
0(0(0(0(x)))) → 0(1(1(1(x))))
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
01(0(0(0(x)))) → 01(1(1(1(x))))
01(0(0(0(x)))) → 11(1(1(x)))
The TRS R consists of the following rules:
0(0(0(0(x)))) → 0(1(1(1(x))))
1(0(1(1(x)))) → 0(0(0(1(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 1 SCC with 1 less node.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
01(0(0(0(x)))) → 01(1(1(1(x))))
The TRS R consists of the following rules:
0(0(0(0(x)))) → 0(1(1(1(x))))
1(0(1(1(x)))) → 0(0(0(1(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
01(0(0(0(x)))) → 01(1(1(1(x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(0(x1)) = | | + | | / | -I | -I | 0A | \ |
| | | 1A | -I | 0A | | |
| \ | 0A | 0A | 0A | / |
| · | x1 |
| POL(1(x1)) = | | + | | / | 0A | -I | 0A | \ |
| | | 1A | 0A | 1A | | |
| \ | -I | -I | 0A | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
1(0(1(1(x)))) → 0(0(0(1(x))))
0(0(0(0(x)))) → 0(1(1(1(x))))
(8) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
0(0(0(0(x)))) → 0(1(1(1(x))))
1(0(1(1(x)))) → 0(0(0(1(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(10) YES