Equivalence relation using tableaux

Question

How can I prove that two formulae are equivalent using analytic tableaux? For example, how can I prove the following theorem? $$ (p \rightarrow q) \equiv (\neg q \rightarrow \neg p)$$

Answer 1

In using tableaux (I've also heard them called truth-trees (as opposed to truth tables)) to prove a sentence $\phi$, we can start with $\lnot\phi$ and show that every branch (a potential way of satisfying $\lnot\phi$) leads to a contradiction. If every potential way of satisfying $\lnot\phi$ is a contradiction, then $\phi$ must be true.

I'd suggest that in using the tableaux for equivalences, you replace $\phi \equiv \psi$ with $(\phi \to \psi) \land (\psi \to \phi)$. I'd also suggest that you allow two “shortcuts” for convenience. The first is that a conditional $\phi \to \psi$ is equivalent to the disjunction $\lnot\phi \lor \psi$, so when a branch has $\phi \to\psi$, it can split with $\lnot\phi$ on one side, and $\psi$ on the other. The second shortcut is that a negated conditional $\lnot(\phi\to\psi)$ is logically equivalent to $\phi \land \lnot\psi$, and this can be expanded in place, too. I will also presume that we can move between $\lnot\lnot\phi$ and $\phi$ without making note of it. Using these shortcuts, we can construct a tree for your formula. This isn't quite the tree presentation that is usually shown (because I don't know how to render that in MathJax), but it should be close enough that you can see the structure.

$$\frac{ \begin{array}{c} \lnot[ (p \to q) \equiv (\lnot q \to \lnot p) ] \\ \lnot[ ((p \to q) \to (\lnot q \to \lnot p)) \land ((\lnot q \to \lnot p) \to (p \to q))] \\ \lnot((p \to q) \to (\lnot q \to \lnot p)) \lor \lnot((\lnot q \to \lnot p) \to (p \to q)) \end{array} }{\displaystyle \frac{ \begin{array}{c} \lnot((p \to q) \to (\lnot q \to \lnot p)) \\ \lnot(\lnot q \to \lnot p) \\ \lnot q \\ p \\ p \to q \end{array}}{ \begin{array}{c} \lnot p \\ \bot \end{array} \qquad \begin{array}{c} q \\ \bot \\ \end{array}} \qquad \frac{ \begin{array}{c} \lnot((\lnot q \to \lnot p) \to (p \to q)) \\ \lnot(p \to q) \\ p \\ \lnot q \\ \lnot q \to \lnot p \end{array} }{ \begin{array}{c} q \\ \bot \end{array} \qquad \begin{array}{c} \lnot p \\ \bot \end{array} } } $$

Since all the branches terminate, $\lnot[ (p \to q) \equiv (\lnot q \to \lnot p) ]$ is unsatisfiable, which means that the following sentence is a theorem.

$$(p \to q) \equiv (\lnot q \to \lnot p)$$

If that notation isn't quite clear, here's a more linear presentation that might help.

1. $\lnot[ (p \to q) \equiv (\lnot q \to \lnot p) ]$
2. $\lnot[ ((p \to q) \to (\lnot q \to \lnot p)) \land ((\lnot q \to \lnot p) \to (p \to q))]$
3. $\lnot((p \to q) \to (\lnot q \to \lnot p)) \lor \lnot((\lnot q \to \lnot p) \to (p \to q))$
4. • Branch for $\lnot((p \to q) \to (\lnot q \to \lnot p))$ from 3.
5. • $(p \to q)$
6. • $\lnot(\lnot q \to \lnot p)$
7. • $\lnot q$
8. • $p$
9. • • Branch for $\lnot p$ from 5.
10. • • $\bot$ from $p, \lnot p$, end branch
11. • • Branch for $q$ from 5.
12. • • $\bot$ from $q, \lnot q$, end branch
13. • Branch for $\lnot((\lnot q \to \lnot p) \to (p \to q))$ from 3.
14. • $\lnot q \to \lnot p$
15. • $\lnot(p \to q)$
16. • $p$
17. • $\lnot q$
18. • • Branch for $q$ from 14.
19. • • $\bot$ from $q, \lnot q$, end branch
20. • • Branch for $\lnot p$ from 14.
21. • • $\bot$ from $p, \lnot p$, end branch

Answer 2

Your class probably defines $A\equiv B$ as $(A\to B)\land (B\to A)$, so you can treat it that way for the tableau.

Answer 3

Is this what you need? I don't understand what you want.