(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(b(a(x)))
c(b(x)) → b(c(c(x)))
Q is empty.
(1) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(b(a(x)))
c(b(x)) → b(c(c(x)))
The set Q consists of the following terms:
a(b(x0))
c(b(x0))
(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(b(x)) → A(x)
C(b(x)) → C(c(x))
C(b(x)) → C(x)
The TRS R consists of the following rules:
a(b(x)) → b(b(a(x)))
c(b(x)) → b(c(c(x)))
The set Q consists of the following terms:
a(b(x0))
c(b(x0))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(b(x)) → C(x)
C(b(x)) → C(c(x))
The TRS R consists of the following rules:
a(b(x)) → b(b(a(x)))
c(b(x)) → b(c(c(x)))
The set Q consists of the following terms:
a(b(x0))
c(b(x0))
We have to consider all minimal (P,Q,R)-chains.
(8) 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.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(b(x)) → C(x)
C(b(x)) → C(c(x))
The TRS R consists of the following rules:
c(b(x)) → b(c(c(x)))
The set Q consists of the following terms:
a(b(x0))
c(b(x0))
We have to consider all minimal (P,Q,R)-chains.
(10) 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].
a(b(x0))
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(b(x)) → C(x)
C(b(x)) → C(c(x))
The TRS R consists of the following rules:
c(b(x)) → b(c(c(x)))
The set Q consists of the following terms:
c(b(x0))
We have to consider all minimal (P,Q,R)-chains.
(12) 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(b(x)) → C(x)
C(b(x)) → C(c(x))
Used ordering: Polynomial interpretation [POLO]:
POL(C(x1)) = 3·x1
POL(b(x1)) = 1 + x1
POL(c(x1)) = x1
(13) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
c(b(x)) → b(c(c(x)))
The set Q consists of the following terms:
c(b(x0))
We have to consider all minimal (P,Q,R)-chains.
(14) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(15) YES
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(x)
The TRS R consists of the following rules:
a(b(x)) → b(b(a(x)))
c(b(x)) → b(c(c(x)))
The set Q consists of the following terms:
a(b(x0))
c(b(x0))
We have to consider all minimal (P,Q,R)-chains.
(17) 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.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(x)
R is empty.
The set Q consists of the following terms:
a(b(x0))
c(b(x0))
We have to consider all minimal (P,Q,R)-chains.
(19) 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].
a(b(x0))
c(b(x0))
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- A(b(x)) → A(x)
The graph contains the following edges 1 > 1
(22) YES