dear all

How can I prove Qa: p(X,Y) is not contained in Qb: p(A,'a')? I have
proved the containment in the other direction using mapping, but not
sure how to proceed with this case.


Thanks in advance
Helen

On Feb 10, 5:18*pm, Helen2007 <helen.ack... (AT) googlemail (DOT) com> wrote:
I don't know this notation or terms. One can guess that you want to
prove
~(forall a [p(a, 'a')] -> forall x, y [p(x, y)])
Suggest you try a counterexample.

Googling +"Qa: p(X,Y)" +"contained in" gives a single hit 'Logical
query optimization by proof-tree transformation' (Ramakrishnan/Sagiv/
Ullman/Vardi 1993) "... have frontiers that are related to the p(X, Y)
nodes by Qa: p(X, Y) : a(X, Z) p(Z, ... A set of proof trees S is said
to be contained in another, say, T, if every fact ...", consistent
with my guess. Context is helpful for any question, like the topic or
paper or textbook you're reading & it's polite when it's homework to
say so.

philip