(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
0(P(x)) → 1(P(x))
1(P(x)) → c(P(x))
0(c(x)) → 1(0(x))
1(c(x)) → c(0(x))
Q is empty.
(1) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
0(P(x)) → 1(P(x))
1(P(x)) → c(P(x))
0(c(x)) → 1(0(x))
1(c(x)) → c(0(x))
The set Q consists of the following terms:
0(P(x0))
1(P(x0))
0(c(x0))
1(c(x0))
(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:
01(P(x)) → 11(P(x))
01(c(x)) → 11(0(x))
01(c(x)) → 01(x)
11(c(x)) → 01(x)
The TRS R consists of the following rules:
0(P(x)) → 1(P(x))
1(P(x)) → c(P(x))
0(c(x)) → 1(0(x))
1(c(x)) → c(0(x))
The set Q consists of the following terms:
0(P(x0))
1(P(x0))
0(c(x0))
1(c(x0))
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(c(x)) → 11(0(x))
11(c(x)) → 01(x)
01(c(x)) → 01(x)
The TRS R consists of the following rules:
0(P(x)) → 1(P(x))
1(P(x)) → c(P(x))
0(c(x)) → 1(0(x))
1(c(x)) → c(0(x))
The set Q consists of the following terms:
0(P(x0))
1(P(x0))
0(c(x0))
1(c(x0))
We have to consider all minimal (P,Q,R)-chains.
(7) 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(c(x)) → 11(0(x))
11(c(x)) → 01(x)
01(c(x)) → 01(x)
Used ordering: Polynomial interpretation [POLO]:
POL(0(x1)) = 2 + 2·x1
POL(01(x1)) = 3 + 3·x1
POL(1(x1)) = 2 + 2·x1
POL(11(x1)) = 2·x1
POL(P(x1)) = x1
POL(c(x1)) = 2 + 2·x1
(8) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
0(P(x)) → 1(P(x))
1(P(x)) → c(P(x))
0(c(x)) → 1(0(x))
1(c(x)) → c(0(x))
The set Q consists of the following terms:
0(P(x0))
1(P(x0))
0(c(x0))
1(c(x0))
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