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