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?

  • $\begingroup$ 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 $\endgroup$ – David Diaz Dec 13 '18 at 20:27
  • $\begingroup$ 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 $ $\endgroup$ – VirtualUser Dec 13 '18 at 20:30
  • 1
    $\begingroup$ One could use the method of analytic tableaux. $\endgroup$ – Shaun Dec 13 '18 at 20:30
  • $\begingroup$ Ok, but my solution is good or not? Thanks for interesting link. $\endgroup$ – VirtualUser Dec 13 '18 at 20:34
  • $\begingroup$ @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. $\endgroup$ – Bram28 Dec 13 '18 at 20:35

You'll eventually get to a correct answer by working out a truth-table (and, as a general method, that is to be preferred over what I'm about to do below), but here is a more direct approach by careful inspection of the statements involved.

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:

$$\alpha = (q \land p \land r) \lor (\neg q \land \neg p \land \neg r)$$

  • $\begingroup$ Thanks for your reply I think that I have understood this problem, it is interesting alternative for table. $\endgroup$ – VirtualUser Dec 13 '18 at 20:59
  • 1
    $\begingroup$ @VirtualUser Yes, the statements were 'nice' enough that I was able to derive it this way pretty quickly ... but the Truth-table method will be more general. $\endgroup$ – Bram28 Dec 13 '18 at 21:13

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.