# What is the significance of conditional tautology

In your approach, you made a mistake in the derivation on the third line, it should be: $$[\lnot (\lnot p \lor q) \lor \lnot (q \rightarrow r)] \lor (p \rightarrow r)$$

so that your fourth line becomes:

$$[(p\land \lnot q) \lor (q \land \lnot r)] \lor (p \rightarrow r)$$

At this point, there isn't really much to be done except for a tedious case-by-case argument ("Suppose $p$ is true. *[...]* Now suppose $p$ is false *[...]*.").

Let us take a step back. When proving a tautology (which is not easily seen to be so by virtue of known theorems), it is often a more productive technique to try and prove that its negation is a contradiction. The idea behind this is that if the original $\phi$ is to be a tautology, then by investigating its negation $\neg \phi$, we will obtain stricter and stricter conditions on $p, q, r$ to allow $\neg \phi$ to be true, until we get to the conclusion that there are *no* possibilities left, i.e. that $\phi$ is a tautology.

Essentially, one converts a universal check ("All interpretations yield true") to a search for an example ("Some interpretation does not yield true") -- keeping in mind that we want the latter search to be fruitless. Basically, it's easier to find the black sheep in a flock than to count the white ones. For a more detailed account of this approach, you can search for terms like *semantic tableaux* or *analytic tableaux*.

Returning to our concrete example, let us see how this approach works out. We use repeatedly the equivalences $p \to q \iff \neg p \lor q$ and $\neg (p \to q) \iff p \land \neg q$:

\begin{align*} & \neg\big([(p\rightarrow q) \land (q \rightarrow r)] \rightarrow (p \rightarrow r )\big)\\ \iff& (p\to q)\land (q\to r) \land \neg(p \to r) \\ \iff& (\neg p \lor q) \land (\neg q \lor r) \land p \land \neg r \end{align*}

From the last two clauses, we see that $p$ must be true and $r$ must be false. But then for $\neg p \lor q$ to be true, $q$ must be true as well. Similarly, $\neg q$ must be true in order for $\neg q \lor r$ to be true as well. Hence we need $q$ to be both true and false at the same time, a contradiction.

We conclude that $[(p\rightarrow q) \land (q \rightarrow r)] \rightarrow (p \rightarrow r )$ is a tautology, as desired.

answered Jan 24 '15 at 22:56

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