(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(d(d(x)))) → d(d(b(b(x))))
a(a(x)) → b(b(b(b(b(b(x))))))
b(b(d(d(b(b(x)))))) → a(a(c(c(x))))
c(c(x)) → d(d(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:
A(a(d(d(x)))) → B(b(x))
A(a(d(d(x)))) → B(x)
A(a(x)) → B(b(b(b(b(b(x))))))
A(a(x)) → B(b(b(b(b(x)))))
A(a(x)) → B(b(b(b(x))))
A(a(x)) → B(b(b(x)))
A(a(x)) → B(b(x))
A(a(x)) → B(x)
B(b(d(d(b(b(x)))))) → A(a(c(c(x))))
B(b(d(d(b(b(x)))))) → A(c(c(x)))
B(b(d(d(b(b(x)))))) → C(c(x))
B(b(d(d(b(b(x)))))) → C(x)
The TRS R consists of the following rules:
a(a(d(d(x)))) → d(d(b(b(x))))
a(a(x)) → b(b(b(b(b(b(x))))))
b(b(d(d(b(b(x)))))) → a(a(c(c(x))))
c(c(x)) → d(d(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(b(d(d(b(b(x)))))) → A(a(c(c(x))))
A(a(d(d(x)))) → B(b(x))
B(b(d(d(b(b(x)))))) → A(c(c(x)))
A(a(d(d(x)))) → B(x)
A(a(x)) → B(b(b(b(b(b(x))))))
A(a(x)) → B(b(b(b(b(x)))))
A(a(x)) → B(b(b(b(x))))
A(a(x)) → B(b(b(x)))
A(a(x)) → B(b(x))
A(a(x)) → B(x)
The TRS R consists of the following rules:
a(a(d(d(x)))) → d(d(b(b(x))))
a(a(x)) → b(b(b(b(b(b(x))))))
b(b(d(d(b(b(x)))))) → a(a(c(c(x))))
c(c(x)) → d(d(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(d(d(x)))) → B(b(x))
A(a(d(d(x)))) → B(x)
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = 2·x1
POL(B(x1)) = 2·x1
POL(a(x1)) = x1
POL(b(x1)) = x1
POL(c(x1)) = 1 + 2·x1
POL(d(x1)) = 1 + 2·x1
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(b(d(d(b(b(x)))))) → A(a(c(c(x))))
B(b(d(d(b(b(x)))))) → A(c(c(x)))
A(a(x)) → B(b(b(b(b(b(x))))))
A(a(x)) → B(b(b(b(b(x)))))
A(a(x)) → B(b(b(b(x))))
A(a(x)) → B(b(b(x)))
A(a(x)) → B(b(x))
A(a(x)) → B(x)
The TRS R consists of the following rules:
a(a(d(d(x)))) → d(d(b(b(x))))
a(a(x)) → b(b(b(b(b(b(x))))))
b(b(d(d(b(b(x)))))) → a(a(c(c(x))))
c(c(x)) → d(d(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
B(b(d(d(b(b(x)))))) → A(c(c(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = x1
POL(B(x1)) = 1
POL(a(x1)) = 1
POL(b(x1)) = 1
POL(c(x1)) = 0
POL(d(x1)) = 0
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
c(c(x)) → d(d(x))
a(a(d(d(x)))) → d(d(b(b(x))))
a(a(x)) → b(b(b(b(b(b(x))))))
b(b(d(d(b(b(x)))))) → a(a(c(c(x))))
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(b(d(d(b(b(x)))))) → A(a(c(c(x))))
A(a(x)) → B(b(b(b(b(b(x))))))
A(a(x)) → B(b(b(b(b(x)))))
A(a(x)) → B(b(b(b(x))))
A(a(x)) → B(b(b(x)))
A(a(x)) → B(b(x))
A(a(x)) → B(x)
The TRS R consists of the following rules:
a(a(d(d(x)))) → d(d(b(b(x))))
a(a(x)) → b(b(b(b(b(b(x))))))
b(b(d(d(b(b(x)))))) → a(a(c(c(x))))
c(c(x)) → d(d(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
A(a(x)) → B(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( A(x1) ) = max{0, 2x1 - 1} |
| POL( B(x1) ) = max{0, x1 - 1} |
| POL( d(x1) ) = max{0, -2} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
c(c(x)) → d(d(x))
a(a(d(d(x)))) → d(d(b(b(x))))
a(a(x)) → b(b(b(b(b(b(x))))))
b(b(d(d(b(b(x)))))) → a(a(c(c(x))))
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(b(d(d(b(b(x)))))) → A(a(c(c(x))))
A(a(x)) → B(b(b(b(b(b(x))))))
A(a(x)) → B(b(b(b(b(x)))))
A(a(x)) → B(b(b(b(x))))
A(a(x)) → B(b(b(x)))
A(a(x)) → B(b(x))
The TRS R consists of the following rules:
a(a(d(d(x)))) → d(d(b(b(x))))
a(a(x)) → b(b(b(b(b(b(x))))))
b(b(d(d(b(b(x)))))) → a(a(c(c(x))))
c(c(x)) → d(d(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.
B(b(d(d(b(b(x)))))) → A(a(c(c(x))))
A(a(x)) → B(b(b(b(b(x)))))
A(a(x)) → B(b(b(b(x))))
A(a(x)) → B(b(b(x)))
A(a(x)) → B(b(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = 2·x1
POL(B(x1)) = 2·x1
POL(a(x1)) = 5 + x1
POL(b(x1)) = 1 + x1
POL(c(x1)) = 2·x1
POL(d(x1)) = 2·x1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
c(c(x)) → d(d(x))
a(a(d(d(x)))) → d(d(b(b(x))))
a(a(x)) → b(b(b(b(b(b(x))))))
b(b(d(d(b(b(x)))))) → a(a(c(c(x))))
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(a(x)) → B(b(b(b(b(b(x))))))
The TRS R consists of the following rules:
a(a(d(d(x)))) → d(d(b(b(x))))
a(a(x)) → b(b(b(b(b(b(x))))))
b(b(d(d(b(b(x)))))) → a(a(c(c(x))))
c(c(x)) → d(d(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