(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
R(E(x)) → L(E(x))
a(L(x)) → L(Aa(x))
b(L(x)) → L(Ab(x))
R(Aa(x)) → a(R(x))
R(Ab(x)) → b(R(x))
a(L(x)) → b(R(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:
E(R(x)) → E(L(x))
L(a(x)) → Aa(L(x))
L(b(x)) → Ab(L(x))
Aa(R(x)) → R(a(x))
Ab(R(x)) → R(b(x))
L(a(x)) → R(b(x))
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(Aa(x1)) = 2 + x1
POL(Ab(x1)) = x1
POL(E(x1)) = x1
POL(L(x1)) = x1
POL(R(x1)) = 1 + x1
POL(a(x1)) = 2 + x1
POL(b(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
E(R(x)) → E(L(x))
L(a(x)) → R(b(x))
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
L(a(x)) → Aa(L(x))
L(b(x)) → Ab(L(x))
Aa(R(x)) → R(a(x))
Ab(R(x)) → R(b(x))
Q is empty.
(5) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
L(a(x)) → Aa(L(x))
L(b(x)) → Ab(L(x))
Aa(R(x)) → R(a(x))
Ab(R(x)) → R(b(x))
The set Q consists of the following terms:
L(a(x0))
L(b(x0))
Aa(R(x0))
Ab(R(x0))
(7) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L1(a(x)) → AA(L(x))
L1(a(x)) → L1(x)
L1(b(x)) → AB(L(x))
L1(b(x)) → L1(x)
The TRS R consists of the following rules:
L(a(x)) → Aa(L(x))
L(b(x)) → Ab(L(x))
Aa(R(x)) → R(a(x))
Ab(R(x)) → R(b(x))
The set Q consists of the following terms:
L(a(x0))
L(b(x0))
Aa(R(x0))
Ab(R(x0))
We have to consider all minimal (P,Q,R)-chains.
(9) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L1(b(x)) → L1(x)
L1(a(x)) → L1(x)
The TRS R consists of the following rules:
L(a(x)) → Aa(L(x))
L(b(x)) → Ab(L(x))
Aa(R(x)) → R(a(x))
Ab(R(x)) → R(b(x))
The set Q consists of the following terms:
L(a(x0))
L(b(x0))
Aa(R(x0))
Ab(R(x0))
We have to consider all minimal (P,Q,R)-chains.
(11) 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.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L1(b(x)) → L1(x)
L1(a(x)) → L1(x)
R is empty.
The set Q consists of the following terms:
L(a(x0))
L(b(x0))
Aa(R(x0))
Ab(R(x0))
We have to consider all minimal (P,Q,R)-chains.
(13) 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(x0))
L(b(x0))
Aa(R(x0))
Ab(R(x0))
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L1(b(x)) → L1(x)
L1(a(x)) → L1(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) 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:
- L1(b(x)) → L1(x)
The graph contains the following edges 1 > 1
- L1(a(x)) → L1(x)
The graph contains the following edges 1 > 1
(16) YES