(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
and(x, or(y, z)) → or(and(x, y), and(x, z))
and(x, and(y, y)) → and(x, y)
or(or(x, y), and(y, z)) → or(x, y)
or(x, and(x, y)) → x
or(true, y) → true
or(x, false) → x
or(x, x) → x
or(x, or(y, y)) → or(x, y)
and(x, true) → x
and(false, y) → false
and(x, x) → 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:
AND(x, or(y, z)) → OR(and(x, y), and(x, z))
AND(x, or(y, z)) → AND(x, y)
AND(x, or(y, z)) → AND(x, z)
AND(x, and(y, y)) → AND(x, y)
OR(x, or(y, y)) → OR(x, y)
The TRS R consists of the following rules:
and(x, or(y, z)) → or(and(x, y), and(x, z))
and(x, and(y, y)) → and(x, y)
or(or(x, y), and(y, z)) → or(x, y)
or(x, and(x, y)) → x
or(true, y) → true
or(x, false) → x
or(x, x) → x
or(x, or(y, y)) → or(x, y)
and(x, true) → x
and(false, y) → false
and(x, x) → 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 1 less node.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
OR(x, or(y, y)) → OR(x, y)
The TRS R consists of the following rules:
and(x, or(y, z)) → or(and(x, y), and(x, z))
and(x, and(y, y)) → and(x, y)
or(or(x, y), and(y, z)) → or(x, y)
or(x, and(x, y)) → x
or(true, y) → true
or(x, false) → x
or(x, x) → x
or(x, or(y, y)) → or(x, y)
and(x, true) → x
and(false, y) → false
and(x, x) → 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:
OR(x, or(y, y)) → OR(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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:
- OR(x, or(y, y)) → OR(x, y)
The graph contains the following edges 1 >= 1, 2 > 2
(9) TRUE
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
AND(x, or(y, z)) → AND(x, z)
AND(x, or(y, z)) → AND(x, y)
AND(x, and(y, y)) → AND(x, y)
The TRS R consists of the following rules:
and(x, or(y, z)) → or(and(x, y), and(x, z))
and(x, and(y, y)) → and(x, y)
or(or(x, y), and(y, z)) → or(x, y)
or(x, and(x, y)) → x
or(true, y) → true
or(x, false) → x
or(x, x) → x
or(x, or(y, y)) → or(x, y)
and(x, true) → x
and(false, y) → false
and(x, x) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) 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.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
AND(x, or(y, z)) → AND(x, z)
AND(x, or(y, z)) → AND(x, y)
AND(x, and(y, y)) → AND(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) 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:
- AND(x, or(y, z)) → AND(x, z)
The graph contains the following edges 1 >= 1, 2 > 2
- AND(x, or(y, z)) → AND(x, y)
The graph contains the following edges 1 >= 1, 2 > 2
- AND(x, and(y, y)) → AND(x, y)
The graph contains the following edges 1 >= 1, 2 > 2
(14) TRUE