# Finding $\alpha$ so that you get two tautologies

I have this task to solve:

Find (if it exists) such a sentence formula α so that the following formulas are simultaneously tautologies of the propositional calculus: $$(q \rightarrow \alpha) \leftrightarrow (q \rightarrow (p \wedge r))$$ and $$(\alpha \rightarrow q) \leftrightarrow (\neg(p \vee r) \rightarrow q)$$

Generally I have idea to solve this task in use of table. So I created a table with 7 columns: $$p$$,$$q$$,$$r$$,$$(p \wedge r)$$, $$(q \rightarrow (p \wedge r)$$ $$(q \rightarrow \alpha)$$, $$LS \leftrightarrow RS$$ and 8 rows with each case. In last column I get these values: $$\alpha$$, $$\neg \alpha$$, $$1$$, $$1$$, $$\neg \alpha$$, $$\neg \alpha$$, $$1$$, $$1$$ So if we consider tautology, both $$\alpha$$ and $$\neg \alpha$$ should be $$1$$ so it can't be done. So $$\alpha$$ like that does not exists and I don't have to consider second potential tautology. Have I right? Eventually, there are other ways to solve tasks like that?

• Have you compared the values of alpha with the values of other columns? Remember, you're looking for a statement or formula, not necessarily a truth value – David Diaz Dec 13 '18 at 20:27
• It should be tautology so $LS \leftrightarrow RS$ must be true value. If I represent this by $\alpha$ it should be true too. The same thing with $\neg \alpha$ – user617243 Dec 13 '18 at 20:30
• One could use the method of analytic tableaux. – Shaun Dec 13 '18 at 20:30
• Ok, but my solution is good or not? Thanks for interesting link. – user617243 Dec 13 '18 at 20:34
• @VirtualUser Your method of creating a truth-table is fine ... but your conclusion that it cannot be done is not. It can be done. Just see what value, if any, is forced on $\alpha$ to make the truth-values come out right in each row. And then just generate an expression for that. – Bram28 Dec 13 '18 at 20:35

First, note that the first conditional holds True whenever $$q$$ is False, so nothing is required of $$\alpha$$ in the case where $$q$$ is False. So the only interesting case is where $$q$$ is True. And in that case, $$\alpha$$ needs to have the same truth-value as $$p \land r$$ in order for the biconditional to be True, i.e. just when $$p$$ and $$r$$ are both True.
Likewise, you can immediately see that the second biconditional is True whenever $$q$$ is True, so here the only interesting case is where $$q$$ is False, and in that case, the biconditional is True just when $$\alpha$$ has the same value as $$\neg (p \lor r)$$, i.e. when both $$p$$ and $$r$$ are False.
So, the only cases where $$\alpha$$ should be True are where $$q$$, $$p$$, and $$r$$ are all True, or when they are all False:
$$\alpha = (q \land p \land r) \lor (\neg q \land \neg p \land \neg r)$$