Predicate logic and validity

Question

I have the following task:

"Show that Pa V Pb -> Ex Px is valid" where E stands for the existential quantifier.

I have done the following: - Let M denote a model with domain D, and assume that M |= Pa V Pb - It suffices then to show that M |= Ex Px - Let s be an element in D, arbitrarily chosen - By assumption, we know that M |= Ps (since we consider disjunction, it suffices to only include one, from Pa V Pb) - Thus, it is the case that M |= Ex Ps - Since s was arbitrarily chosen, it will be so that M |= Ex Px, that Ex Px is true in M.

Is this the right way to prove validity?

Maybe this would also work: from $PaV Pb$ one must be true, say $Pa$. Then I think you can conclude from $Pa$, that $Ex Px$ — user99680, Nov 04 '13 at 10:39

Shaun · Accepted Answer · 2013-11-04T11:19:50.047

0

It looks okay at a glance, but I could be wrong. Using tableux is much more succinct: you assume $Pa\vee Pb$ but also $\neg\exists xPx$ to get

Tab ,

which ends in contradictions. The tableau of the negation of your formula is closed. Hence it is valid.

edited Nov 04 '13 at 11:19

answered Nov 04 '13 at 11:12

Shaun

44,997

1

Thank you! Will look into that. – aprilrocks92 Nov 04 '13 at 11:23
You're welcome. A great place to start is S.G.Simpson's lecture notes on Mathematical Logic. – Shaun Nov 04 '13 at 11:27

Predicate logic and validity

1 Answers1