(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(c(c(b(x)))) → A(1, b(c(x)))
C(c(c(b(x)))) → B(c(x))
C(c(c(b(x)))) → C(x)
B(c(b(c(x)))) → A(0, a(1, x))
B(c(b(c(x)))) → A(1, x)
A(0, x) → C(c(x))
A(0, x) → C(x)
A(1, x) → C(b(x))
A(1, x) → B(x)
The TRS R consists of the following rules:
c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) 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(x)))) → B(c(x))
C(c(c(b(x)))) → C(x)
B(c(b(c(x)))) → A(0, a(1, x))
B(c(b(c(x)))) → A(1, x)
A(0, x) → C(x)
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(1) = 0
POL(A(x1, x2)) = 2 + x1 + 2·x2
POL(B(x1)) = 2 + 2·x1
POL(C(x1)) = 2·x1
POL(a(x1, x2)) = 2 + 2·x1 + x2
POL(b(x1)) = 1 + x1
POL(c(x1)) = 1 + x1
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(c(c(b(x)))) → A(1, b(c(x)))
A(0, x) → C(c(x))
A(1, x) → C(b(x))
A(1, x) → B(x)
The TRS R consists of the following rules:
c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(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:
A(1, x) → C(b(x))
C(c(c(b(x)))) → A(1, b(c(x)))
The TRS R consists of the following rules:
c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) TransformationProof (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
A(
1,
x) →
C(
b(
x)) at position [0] we obtained the following new rules [LPAR04]:
A(1, c(b(c(x0)))) → C(a(0, a(1, x0))) → A(1, c(b(c(x0)))) → C(a(0, a(1, x0)))
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(c(c(b(x)))) → A(1, b(c(x)))
A(1, c(b(c(x0)))) → C(a(0, a(1, x0)))
The TRS R consists of the following rules:
c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) TransformationProof (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
C(
c(
c(
b(
x)))) →
A(
1,
b(
c(
x))) at position [1] we obtained the following new rules [LPAR04]:
C(c(c(b(b(c(x0)))))) → A(1, a(0, a(1, x0))) → C(c(c(b(b(c(x0)))))) → A(1, a(0, a(1, x0)))
C(c(c(b(c(c(b(x0))))))) → A(1, b(a(1, b(c(x0))))) → C(c(c(b(c(c(b(x0))))))) → A(1, b(a(1, b(c(x0)))))
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(1, c(b(c(x0)))) → C(a(0, a(1, x0)))
C(c(c(b(b(c(x0)))))) → A(1, a(0, a(1, x0)))
C(c(c(b(c(c(b(x0))))))) → A(1, b(a(1, b(c(x0)))))
The TRS R consists of the following rules:
c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
A(1, c(b(c(x0)))) → C(a(0, a(1, x0)))
C(c(c(b(b(c(x0)))))) → A(1, a(0, a(1, x0)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :
| POL(A(x1, x2)) = | 2 | + | | · | x1 | + | | · | x2 |
| POL(a(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
a(1, x) → c(b(x))
c(c(c(b(x)))) → a(1, b(c(x)))
a(0, x) → c(c(x))
b(c(b(c(x)))) → a(0, a(1, x))
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(c(c(b(c(c(b(x0))))))) → A(1, b(a(1, b(c(x0)))))
The TRS R consists of the following rules:
c(c(c(b(x)))) → a(1, b(c(x)))
b(c(b(c(x)))) → a(0, a(1, x))
a(0, x) → c(c(x))
a(1, x) → c(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(14) TRUE