(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
c(c(c(a(x)))) → d(d(x))
d(b(x)) → c(c(x))
b(c(x)) → b(a(c(x)))
c(x) → a(a(x))
d(x) → b(c(x))
Q is empty.
(1) QTRS Reverse (EQUIVALENT transformation)
We applied the QTRS Reverse Processor [REVERSE].
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
c(x) → a(a(x))
d(x) → c(b(x))
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(a(x1)) = x1
POL(b(x1)) = 1 + x1
POL(c(x1)) = 2 + x1
POL(d(x1)) = 3 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
c(x) → a(a(x))
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
d(x) → c(b(x))
Q is empty.
(5) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(c(c(c(x)))) → D(d(x))
A(c(c(c(x)))) → D(x)
B(d(x)) → C(c(x))
B(d(x)) → C(x)
C(b(x)) → C(a(b(x)))
C(b(x)) → A(b(x))
D(x) → C(b(x))
D(x) → B(x)
The TRS R consists of the following rules:
a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
d(x) → c(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(x) → C(b(x))
C(b(x)) → A(b(x))
A(c(c(c(x)))) → D(d(x))
D(x) → B(x)
B(d(x)) → C(x)
A(c(c(c(x)))) → D(x)
The TRS R consists of the following rules:
a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
d(x) → c(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) 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(c(c(c(x)))) → D(d(x))
D(x) → B(x)
B(d(x)) → C(x)
A(c(c(c(x)))) → D(x)
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = 2 + x1
POL(B(x1)) = 1 + x1
POL(C(x1)) = 2 + x1
POL(D(x1)) = 3 + x1
POL(a(x1)) = x1
POL(b(x1)) = 1 + x1
POL(c(x1)) = 2 + x1
POL(d(x1)) = 3 + x1
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(x) → C(b(x))
C(b(x)) → A(b(x))
The TRS R consists of the following rules:
a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(b(x)) → c(a(b(x)))
d(x) → c(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
(12) TRUE