# For logical equivalences involving conditionals/biconditionals here, do we need to prove some of them?

So I was wondering for logical equivalences involving conditionals/biconditionals mentioned here, do we need to use some of them as base and prove the rest using them or we just take them all as accepted, for example in case of equivalences involving conditionals if we take $$p\rightarrow q\equiv\neg p\vee q$$ as base we can prove $$p\rightarrow q\equiv\neg q\rightarrow\neg p$$: $$p\rightarrow q\equiv\neg p\vee q$$ $$\equiv q\vee\neg p$$ $$\equiv\neg\neg q\vee\neg p$$ $$\equiv\neg q\rightarrow\neg p$$ and $$p\vee q\equiv\neg p\rightarrow q$$: $$p\vee q\equiv \neg\neg p\vee q$$ $$\equiv \neg p\rightarrow q$$ and $$p\wedge q\equiv\neg(p\rightarrow\neg q)$$: $$p\wedge q\equiv\neg\neg p\wedge\neg\neg q$$ $$\equiv\neg(\neg p\vee\neg q)$$ $$\equiv\neg(p\rightarrow\neg q)$$ and in case of equivalences involving biconditionals if we take $$p\leftrightarrow q\equiv(p\rightarrow q)\wedge(q\rightarrow p)$$ and $$p\rightarrow q\equiv\neg p\vee q$$ as base we can prove $$p\leftrightarrow q\equiv\neg p\leftrightarrow\neg q$$: $$p\leftrightarrow q\equiv(p\rightarrow q)\wedge(q\rightarrow p)$$ $$\equiv(\neg p\vee q)\wedge(\neg q\vee p)$$ $$\equiv(q\vee\neg p)\wedge(p\vee\neg q)$$ $$\equiv(\neg\neg q\vee\neg p)\wedge(\neg\neg p\vee\neg q)$$ $$\equiv(\neg q\rightarrow\neg p)\wedge(\neg p\rightarrow\neg q)$$ $$\equiv(\neg p\rightarrow\neg q)\wedge(\neg q\rightarrow\neg p)$$ $$\equiv(\neg p\leftrightarrow\neg q)$$

• @Pooria There is of course no law that says which ones we should take as more fundamental ... but you're absolutely right: with the two you indicate, and using the 'standard' equivalence principles involving the 'boolean connectives' $\land$, $\lor$, and $\neg$, we can prove all others, and that's because with these two you can reduce everything to the boolean connectives, and you can show that the 'standard' set is complete in the sense that if $\phi \Leftrightarrow \psi$ then $\phi \dashv \vdash \psi$ ... (continued) – Bram28 Oct 26 '17 at 17:24
• @Pooria ... where $\phi \dashv \vdash \psi$ means that you can transform $\phi$ into $\psi$ (and back of course!) using equivalence principles. Another nice meta-logical result! :) – Bram28 Oct 26 '17 at 17:24
• @Pooria So yes, my personal strategy when dealing with equivalences is indeed to use those two equivalences for the conditional and biconditional to rewrite everything in terms of $\land$, $\lor$, and $\neg$, and go from there .... but it still pays to know a few other principles as well, e.g. contraposition ($p \rightarrow q \Leftrightarrow \neg q \rightarrow \neg p$), Exportation ($p \rightarrow (q \rightarrow r) \Leftrightarrow (p \land q) \rightarrow r$), various Conditional Distributions ($(p \lor q) \rightarrow r \Leftrightarrow (p \rightarrow r) \land (q \rightarrow r)$), and others .. – Bram28 Oct 26 '17 at 17:29
• @Pooria Oh yeah, it is on that list ... does it make sense though? Consider: $P$: John is a male, $Q$: John is unmarried, $R$: John is a bachelor ... use these to interpret the left side and the right side and see what you think ... – Bram28 Oct 26 '17 at 19:54
• @PooriaOh, sorry, no, I was trying to tell you that when it comes to those conditionals and biconditionals, your intuitions may actually sometimes lead you astray. Look again at that very equivalence: the left hand side says that $R$ is true when $P$ and $Q$ are both true. So that could be something like 'if you are both a male and unmarried, then you are a bachelor', which we regard as a true statement ... (continued) – Bram28 Oct 27 '17 at 15:41