(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → d(c(x))
a(b(x)) → c(c(c(x)))
b(b(x)) → a(c(c(x)))
c(c(x)) → b(x)
c(d(x)) → a(a(x))
d(d(x)) → b(a(c(x)))
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(a(x1)) = 15 + x1
POL(b(x1)) = 17 + x1
POL(c(x1)) = 9 + x1
POL(d(x1)) = 21 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
a(b(x)) → c(c(c(x)))
b(b(x)) → a(c(c(x)))
c(c(x)) → b(x)
d(d(x)) → b(a(c(x)))
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → d(c(x))
c(d(x)) → a(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(a(x)) → C(x)
C(d(x)) → A(a(x))
C(d(x)) → A(x)
The TRS R consists of the following rules:
a(a(x)) → d(c(x))
c(d(x)) → a(a(x))
Q is empty.
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:
A(a(x)) → C(x)
C(d(x)) → A(a(x))
C(d(x)) → A(x)
Strictly oriented rules of the TRS R:
a(a(x)) → d(c(x))
c(d(x)) → a(a(x))
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = 2·x1
POL(C(x1)) = 2·x1
POL(a(x1)) = 1 + 2·x1
POL(c(x1)) = 2·x1
POL(d(x1)) = 2 + 2·x1
(6) Obligation:
Q DP problem:
P is empty.
R is empty.
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) YES