(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
b(c(a(x))) → a(b(x))
b(b(b(x))) → c(a(c(x)))
c(d(x)) → d(c(x))
c(d(b(x))) → d(c(c(x)))
d(c(x)) → b(b(b(x)))
c(b(x)) → d(a(x))
d(b(c(x))) → a(a(x))
d(a(x)) → b(x)
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(a(x1)) = x1
POL(b(x1)) = 1 + x1
POL(c(x1)) = 1 + x1
POL(d(x1)) = 2 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
b(c(a(x))) → a(b(x))
b(b(b(x))) → c(a(c(x)))
d(b(c(x))) → a(a(x))
d(a(x)) → b(x)
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
c(d(x)) → d(c(x))
c(d(b(x))) → d(c(c(x)))
d(c(x)) → b(b(b(x)))
c(b(x)) → d(a(x))
Q is empty.
(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:
C(d(x)) → D(c(x))
C(d(x)) → C(x)
C(d(b(x))) → D(c(c(x)))
C(d(b(x))) → C(c(x))
C(d(b(x))) → C(x)
C(b(x)) → D(a(x))
The TRS R consists of the following rules:
c(d(x)) → d(c(x))
c(d(b(x))) → d(c(c(x)))
d(c(x)) → b(b(b(x)))
c(b(x)) → d(a(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 3 less nodes.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(d(b(x))) → C(c(x))
C(d(x)) → C(x)
C(d(b(x))) → C(x)
The TRS R consists of the following rules:
c(d(x)) → d(c(x))
c(d(b(x))) → d(c(c(x)))
d(c(x)) → b(b(b(x)))
c(b(x)) → d(a(x))
Q is empty.
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:
C(d(b(x))) → C(c(x))
C(d(x)) → C(x)
C(d(b(x))) → C(x)
Used ordering: Polynomial interpretation [POLO]:
POL(C(x1)) = 2·x1
POL(a(x1)) = x1
POL(b(x1)) = 1 + x1
POL(c(x1)) = 1 + x1
POL(d(x1)) = 2 + x1
(8) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
c(d(x)) → d(c(x))
c(d(b(x))) → d(c(c(x)))
d(c(x)) → b(b(b(x)))
c(b(x)) → d(a(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