(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
2(1(x)) → 4(1(x))
3(4(x)) → 4(2(x))
2(4(x)) → 3(2(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:
2(1(x)) → 4(1(x))
3(4(x)) → 4(2(x))
2(4(x)) → 3(2(x))
The set Q consists of the following terms:
2(1(x0))
3(4(x0))
2(4(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:
31(4(x)) → 21(x)
21(4(x)) → 31(2(x))
21(4(x)) → 21(x)
The TRS R consists of the following rules:
2(1(x)) → 4(1(x))
3(4(x)) → 4(2(x))
2(4(x)) → 3(2(x))
The set Q consists of the following terms:
2(1(x0))
3(4(x0))
2(4(x0))
We have to consider all minimal (P,Q,R)-chains.
(5) 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:
31(4(x)) → 21(x)
21(4(x)) → 31(2(x))
21(4(x)) → 21(x)
Used ordering: Polynomial interpretation [POLO]:
POL(1(x1)) = 2·x1
POL(2(x1)) = 2 + 2·x1
POL(21(x1)) = 1 + 2·x1
POL(3(x1)) = 2 + 2·x1
POL(31(x1)) = x1
POL(4(x1)) = 2 + 2·x1
(6) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
2(1(x)) → 4(1(x))
3(4(x)) → 4(2(x))
2(4(x)) → 3(2(x))
The set Q consists of the following terms:
2(1(x0))
3(4(x0))
2(4(x0))
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