(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(a(x))) → a(b(c(x)))
c(a(x)) → a(c(x))
c(b(x)) → b(a(x))
L(a(a(x))) → L(a(b(c(x))))
c(R(x)) → b(a(R(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:
B(a(a(x))) → B(c(x))
B(a(a(x))) → C(x)
C(a(x)) → C(x)
C(b(x)) → B(a(x))
L1(a(a(x))) → L1(a(b(c(x))))
L1(a(a(x))) → B(c(x))
L1(a(a(x))) → C(x)
C(R(x)) → B(a(R(x)))
The TRS R consists of the following rules:
b(a(a(x))) → a(b(c(x)))
c(a(x)) → a(c(x))
c(b(x)) → b(a(x))
L(a(a(x))) → L(a(b(c(x))))
c(R(x)) → b(a(R(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 2 SCCs with 3 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(a(x))) → C(x)
C(a(x)) → C(x)
C(b(x)) → B(a(x))
B(a(a(x))) → B(c(x))
The TRS R consists of the following rules:
b(a(a(x))) → a(b(c(x)))
c(a(x)) → a(c(x))
c(b(x)) → b(a(x))
L(a(a(x))) → L(a(b(c(x))))
c(R(x)) → b(a(R(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(a(x))) → C(x)
C(a(x)) → C(x)
C(b(x)) → B(a(x))
B(a(a(x))) → B(c(x))
The TRS R consists of the following rules:
c(a(x)) → a(c(x))
c(b(x)) → b(a(x))
c(R(x)) → b(a(R(x)))
b(a(a(x))) → a(b(c(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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:
B(a(a(x))) → C(x)
C(a(x)) → C(x)
B(a(a(x))) → B(c(x))
Used ordering: Polynomial interpretation [POLO]:
POL(B(x1)) = x1
POL(C(x1)) = 2 + 2·x1
POL(R(x1)) = x1
POL(a(x1)) = 2 + 2·x1
POL(b(x1)) = x1
POL(c(x1)) = 2 + 2·x1
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(b(x)) → B(a(x))
The TRS R consists of the following rules:
c(a(x)) → a(c(x))
c(b(x)) → b(a(x))
c(R(x)) → b(a(R(x)))
b(a(a(x))) → a(b(c(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(11) TRUE
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L1(a(a(x))) → L1(a(b(c(x))))
The TRS R consists of the following rules:
b(a(a(x))) → a(b(c(x)))
c(a(x)) → a(c(x))
c(b(x)) → b(a(x))
L(a(a(x))) → L(a(b(c(x))))
c(R(x)) → b(a(R(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L1(a(a(x))) → L1(a(b(c(x))))
The TRS R consists of the following rules:
b(a(a(x))) → a(b(c(x)))
c(a(x)) → a(c(x))
c(b(x)) → b(a(x))
L(a(a(x))) → L(a(b(c(x))))
c(R(x)) → b(a(R(x)))
The set Q consists of the following terms:
b(a(a(x0)))
c(a(x0))
c(b(x0))
L(a(a(x0)))
c(R(x0))
We have to consider all minimal (P,Q,R)-chains.
(15) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L1(a(a(x))) → L1(a(b(c(x))))
The TRS R consists of the following rules:
c(a(x)) → a(c(x))
c(b(x)) → b(a(x))
c(R(x)) → b(a(R(x)))
b(a(a(x))) → a(b(c(x)))
The set Q consists of the following terms:
b(a(a(x0)))
c(a(x0))
c(b(x0))
L(a(a(x0)))
c(R(x0))
We have to consider all minimal (P,Q,R)-chains.
(17) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
L(a(a(x0)))
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L1(a(a(x))) → L1(a(b(c(x))))
The TRS R consists of the following rules:
c(a(x)) → a(c(x))
c(b(x)) → b(a(x))
c(R(x)) → b(a(R(x)))
b(a(a(x))) → a(b(c(x)))
The set Q consists of the following terms:
b(a(a(x0)))
c(a(x0))
c(b(x0))
c(R(x0))
We have to consider all minimal (P,Q,R)-chains.
(19) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
L1(a(a(x))) → L1(a(b(c(x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( b(x1) ) = max{0, x1 - 1} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
c(a(x)) → a(c(x))
c(b(x)) → b(a(x))
c(R(x)) → b(a(R(x)))
b(a(a(x))) → a(b(c(x)))
(20) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
c(a(x)) → a(c(x))
c(b(x)) → b(a(x))
c(R(x)) → b(a(R(x)))
b(a(a(x))) → a(b(c(x)))
The set Q consists of the following terms:
b(a(a(x0)))
c(a(x0))
c(b(x0))
c(R(x0))
We have to consider all minimal (P,Q,R)-chains.
(21) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(22) YES