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