(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
1(0(x)) → 0(0(0(1(x))))
0(1(x)) → 1(x)
1(1(x)) → 0(0(0(0(x))))
0(0(x)) → 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:
11(0(x)) → 01(0(0(1(x))))
11(0(x)) → 01(0(1(x)))
11(0(x)) → 01(1(x))
11(0(x)) → 11(x)
11(1(x)) → 01(0(0(0(x))))
11(1(x)) → 01(0(0(x)))
11(1(x)) → 01(0(x))
11(1(x)) → 01(x)
The TRS R consists of the following rules:
1(0(x)) → 0(0(0(1(x))))
0(1(x)) → 1(x)
1(1(x)) → 0(0(0(0(x))))
0(0(x)) → 0(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 7 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
11(0(x)) → 11(x)
The TRS R consists of the following rules:
1(0(x)) → 0(0(0(1(x))))
0(1(x)) → 1(x)
1(1(x)) → 0(0(0(0(x))))
0(0(x)) → 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].
The following pairs can be oriented strictly and are deleted.
11(0(x)) → 11(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
11(
x1) =
x1
0(
x1) =
0(
x1)
1(
x1) =
1(
x1)
Recursive Path Order [RPO].
Precedence:
11 > 01
The following usable rules [FROCOS05] were oriented:
1(0(x)) → 0(0(0(1(x))))
0(1(x)) → 1(x)
1(1(x)) → 0(0(0(0(x))))
0(0(x)) → 0(x)
(6) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
1(0(x)) → 0(0(0(1(x))))
0(1(x)) → 1(x)
1(1(x)) → 0(0(0(0(x))))
0(0(x)) → 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) TRUE