True or False: if $\alpha \vDash (\beta\land\gamma)$ then $\alpha \vDash \beta$ and $\alpha \vDash \gamma$.

Answer: true because for the and statement to be correct both $\beta$ and $\gamma$ would have to be the same.

True or False: if $\alpha \vDash (\beta\lor\gamma)$ then $\alpha \vDash \beta$ or $\alpha \vDash \gamma$ (or both)

For this one I am unsure. any insight?

  • $\begingroup$ In the first, $\beta$ and $\gamma$ would not have to be the same. Just taking any two distinct things provable from $\alpha$; these won't be the same by our choice, but $\alpha$ certainly proves their conjugation. Do you have the Completeness Theorem at your disposal? $\endgroup$ – Hayden Mar 25 '15 at 0:56

The first one is indeed true, but not for the reason you give. For example, if we set $\alpha \equiv Q \land P$, $\beta\equiv P$, and $\gamma\equiv Q$ (where $P$ and $Q$ are propositional variables), then the premise becomes $$ Q\land P \vDash P\land Q$$ which certainly is the case, even though $P$ and $Q$ are not the same.

Rather your argument should be that for if we have a valuation/interpretation that satisfies $\beta\land\gamma$, then by definition of $\vDash$ it also has to satisfy $\beta$ and $\gamma$ separately.

In the second question, consider what happens if we set $\alpha \equiv P \lor Q$, $\beta\equiv P$, and $\gamma\equiv Q$? Does the assumption hold? Does the purported conclusion?

| cite | improve this answer | |
  • $\begingroup$ yes it should hold as well then. if we have an interpetation that satisifes b V y then by definition it has to satisfy b or y, or can satisfy both. is my thinking sound? $\endgroup$ – user2510809 Mar 25 '15 at 1:08
  • 2
    $\begingroup$ @user251080: No. The question doesn't ask about a single interpretation. Find your definition of $\alpha\vDash\beta$ -- it speaks about all interpretations that satisfy $\alpha$. So in order for the second claim to be true it must either be the case that all interpretations that satisfy $\alpha$ also satisfy $\beta$, or that all these interpretations also satisfy $\gamma$. It won't work that some of them satisfy $\beta$ and others satisfy $\gamma$. $\endgroup$ – hmakholm left over Monica Mar 25 '15 at 1:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.