How to prove that $P \rightarrow Q$ is equivalent with $\neg P \lor Q $?

Question

In my book about Logic, which is called 'Language, Proof and Logic', by the way, there is explained that the conditional $ P \rightarrow Q $ is equivalent with $\neg P \lor Q$.

There is another answer on math.stackexchange that gives as answer: 'just compare the truth tables and you can see that they are equivalent'. However, I would like to know how to prove this using formal propositional logic. This should be possible, right?

Which kind of formal proof system for propositional logic do you want to prove the equivalence in? There are several quite different ones. — hmakholm left over Monica, Sep 17 '14 at 22:35
This is often given as the definition, although I prefer $P\implies Q\equiv \neg[P \land \neg Q]$. — Dan Christensen, Sep 18 '14 at 17:27
Possible duplicate of Equivalence of $a \rightarrow b$ and $\lnot a \vee b$ — , Nov 13 '16 at 12:22

Git Gud · Accepted Answer · 2014-09-19T10:03:04.853

Layout:

To prove that $\neg P\lor Q$ is a formal consequence of $P\to Q$, start by assuming $P\to Q$ and further suppose that $\neg (\neg P\lor Q)$ holds. At this point you should prove $P\lor \neg P$ and perform $\lor$-$\text{Elim}$ on this disjunction. It's easy to find contradictions on both cases yielding $\neg \neg (\neg P\lor Q)$.

For the other direction, naturally start by assuming that $\neg P\lor Q$ holds and then assume $P$. Now start yet another subproof (within the assumption that $P$ holds) assuming that $\neg Q$ holds. At this point perform $\lor$-$\text{Elim}$ on the assumption $\neg P\lor Q$. It's easy to find a contradiction in this last subproof.

score 4 · Answer 2 · edited Jan 05 '20 at 06:08

4

$P → Q$ (given)
$Q \lor \neg Q$ (tautology)
$\neg Q → \neg P$ (contrapositive of (1) )
from (2) and (3): $Q \lor \neg P$ (modus ponens)

edited Jan 05 '20 at 06:08

statcks

15

answered Aug 15 '16 at 17:39

Καrτhικ

141

score 0 · Answer 3 · answered Jan 05 '20 at 15:33

You may translate the two expressions using sheffer's stroke, symbol : |

X | Y means , by definition ~ ( X & Y)
X --> Y means ~ ( X & ~ Y) and, hence : X | ~Y
X v Y means ~ ( ~X & ~Y) , hence : ~X | ~Y

Now, let ~P play the role of X and Q play the role of Y.

~P v Q means (~ ~ P | ~Q), that is : ( P | ~Q) , using double negation.

Let P play the role of X and Q play the role of Y.

( P --> Q) means ( P | ~Q).

The sheffer's stroke translation shows that the two formulas mean exactly the same thing; hence they are equivalent.

score 0 · Answer 4 · answered Jun 15 '24 at 15:45

Alternatively, one would show both implications.

Showing that $(P \implies Q) \implies (\neg P \vee Q)$: So we assume $(P \implies Q)$ is true. Then if $(\neg P \vee Q)$ is false, we get a contradiction. If $(P \implies Q)$ is true, we can't have $P$ true and $Q$ false, which is the only alternative rendering $\neg P \vee Q$ false. So $\neg P \vee Q$ must be true in that case. Thus, the one-way implication holds.

For the other direction, assume $\neg P \vee Q$ is true. $P \implies Q$ is only false in the case $P = \top$ and $Q = \bot$, which makes $\neg P \vee Q$ false, a contradiction.

How to prove that $P \rightarrow Q$ is equivalent with $\neg P \lor Q $?

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