(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(b(L(x))) → b(a(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(b(a(x))) → R(a(b(x)))
Q is empty.
(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:
E1(R(x)) → E1(L(x))
E1(R(x)) → L1(x)
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:
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(b(a(x))) → R(a(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 2 SCCs with 3 less nodes.
(6) Complex Obligation (AND)
(7) 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:
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(b(a(x))) → R(a(b(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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.
(9) 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.
(10) 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
(11) YES
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
E1(R(x)) → E1(L(x))
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(b(a(x))) → R(a(b(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) 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.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
E1(R(x)) → E1(L(x))
The TRS R consists of the following rules:
L(a(x)) → Aa(L(x))
L(b(x)) → Ab(L(x))
L(b(a(x))) → R(a(b(x)))
Ab(R(x)) → R(b(x))
Aa(R(x)) → R(a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) 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:
E1(R(x)) → E1(L(x))
Strictly oriented rules of the TRS R:
L(b(a(x))) → R(a(b(x)))
Ab(R(x)) → R(b(x))
Used ordering: Polynomial interpretation [POLO]:
POL(Aa(x1)) = 2 + x1
POL(Ab(x1)) = 2·x1
POL(E1(x1)) = 3·x1
POL(L(x1)) = 2·x1
POL(R(x1)) = 1 + 2·x1
POL(a(x1)) = 1 + x1
POL(b(x1)) = 2·x1
(16) Obligation:
Q DP problem:
P is empty.
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))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(18) YES