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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 $p\leftrightarrow q\equiv\neg p\leftrightarrow\neg q$: $$p\leftrightarrow q\equiv(p\rightarrow q)\wedge(q\rightarrow p)$$ $$\equiv(\neg p\vee q)\wedge(\neg q\vee p)$$ $$\equiv(q\vee\neg p)\wedge(p\vee\neg q)$$ $$\equiv(\neg\neg q\vee\neg p)\wedge(\neg\neg p\vee\neg q)$$ $$\equiv(\neg q\rightarrow\neg p)\wedge(\neg p\rightarrow\neg q)$$ $$\equiv(\neg p\rightarrow\neg q)\wedge(\neg q\rightarrow\neg p)$$ $$\equiv(\neg p\leftrightarrow\neg q)$$

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Pooria
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As illustrated in your very question, you can indeed derive further equivalences from more 'basic' equivalences. However, at some point we will need to establish those basic equivalences themselves. That is, we cannot, as you put it, "just take them all as accepted", and in fact we cannnot "just" take any equivalence as accepted.

Every equivalence needs to be justified, and we either derive them from earlier established/justified equivalences, or we justify them on the basis of the formal semantics of how these truth-functions are defined (or, a little less formally, but still amounting to the same thing, on the basis of truth-tables).

Bram28
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  • Glad to see the legendary teacher here ^^ , So which ones should we take as base and start achieving the rest from there? I've noticed if we take $p\rightarrow q\equiv\neg p\vee q$ and $p\leftrightarrow q\equiv(p\rightarrow q)\wedge(q\rightarrow p)$ all the rest can be achieved. – Pooria Oct 26 '17 at 16:59
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    @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
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    @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
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    @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
  • So I shouldn't bother taking those two as base and proving the rest based on them because there's no rule saying those two are fundamental! and we're free to use any of the equivalences in the list of equivalences involving conditionals/biconditionals whenever needed. – Pooria Oct 26 '17 at 18:26
  • @Pooria Once they're proven, go ahead and use them! But if you ever run into a new one, make sure to check it's really valid before using it. Do you think the following is valid? $(P \land Q) \rightarrow R \Leftrightarrow (P \rightarrow R) \lor (Q \rightarrow R)$ – Bram28 Oct 26 '17 at 19:38
  • Well if they weren't proven they wouldn't be on that list on that page :D, I just wonder how they have been proven, as with that following one it's on that list too and perfectly valid. – Pooria Oct 26 '17 at 19:48
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    @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
  • Hmmm I think I get it, that's a tautology so $(P \wedge Q) \rightarrow R$ is equivalent to $(P \rightarrow R) \vee (Q \rightarrow R)$ – Pooria Oct 26 '17 at 20:20
  • @Pooria yes, it is a truth-functional tautology ... but does it make intuitive sense that it is? – Bram28 Oct 26 '17 at 22:18
  • Yup, I think you're trying to tell me that the equivalences on that list have been achieved independent of each other, I was thinking it has to be this way that some of them were established first and the rest later using those. – Pooria Oct 27 '17 at 11:37
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    @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
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    ... The right side, on the other hand, says that $R$ is true when $P$ is true, or that $R$ is true when $Q$ is true, i.e that at least one of the following claims is true 'if you are a male, then you are a bachelor' and 'if you are unmarried then you are a bachelor' ... but both of those statements we regard as false! So what is going on?! Well, this is an example of the Paradox of Material Implication, which arises from the discrepancy between the way we use and think about conditionals in real life, and how conditionals are mathematically defined as truth-functions in math and logic. – Bram28 Oct 27 '17 at 15:44
  • BTW I'm curious how has someone come up with those equivalences on those two lists(conditionals+biconditionals)? is it that they have first established some and then the rest using them or they have established each of them independent of others? – Pooria Oct 27 '17 at 16:23
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    @Pooria Well, the funny thing is that several of them are in accordance to our intuitions. So, probably someone was looking at some intuitive way we relate the conditionals and biconditionals, and how they relate to the other connectives, and then verified them using truth-tables. – Bram28 Oct 27 '17 at 16:28