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