The biconditional operator $\iff$ of Propositional Logic can be defined by the identity

$p \iff q \equiv (\lnot p \lor q) \land (\lnot q \lor p) \quad (1.1)$

Use the identity $(1.1)$ and identities from the list on page 2, to show algebraically that

$p \iff q \equiv (\lnot p \land \lnot q) \lor (q \land p)$

State which identity you are using at each step.

What does this question mean by asking to "show algebraically"? I have tried referring to my notes and online search but no luck with a definition! These propositions are already algebraic, are they not? Having some issues understanding the wording of the question. This is from a mock university test. I'm assuming it wants me to demonstrate using a truth table?

  • 1
    $\begingroup$ "To show algebraically" means to demonstrate on the basis of symbolic rules. I encourage you to edit your question adding the identities the problem statement refers to. $\endgroup$
    – Acsor
    Commented Aug 14, 2020 at 15:15

1 Answer 1


Although you haven't posted any of the logical rules the problem statement refers to, it is reasonable to assume they are similar to those found in many standard logic textbooks. To prove your identity, I will make use of

$$ \begin{align} P \land Q & \equiv Q \land P, &\text{$\land$ commutativity} \\ P \land (Q \lor R) & \equiv (P \land Q) \lor (P \land R), &\text{$\land$ distribution} \\ P \land \lnot P & \equiv \bot &\text {(3)} \\ P \lor \bot &\equiv P, & \text{$\bot$ annihilation} \end{align} $$

where $\bot$ denotes falsehood. With that in mind

$$ \begin{align} p \iff q \equiv & \\ (\lnot p \lor q) \land (\lnot q \lor p) \equiv & & \text{By definition}\\ ((\lnot p \lor q) \land \lnot q) \lor ((\lnot p \lor q) \land p) \equiv & &\text{By distribution of $\land$} \\ (\lnot q \land (\lnot p \lor q)) \lor (p \land (\lnot p \lor q)) \equiv & &\text{By commutativity of $\land$} \\ (\lnot q \land \lnot p \lor \lnot q \land q) \lor (p \land \lnot p \lor p \land q) \equiv & &\text{By distribution of $\land$} \\ (\lnot q \land \lnot p \lor \bot) \lor (\bot \lor p \land q) \equiv & &\text{By $(3)$} \\ (\lnot q \land \lnot p) \lor (p \land q) \equiv & &\text{By $\bot$ annihilation} \\ (\lnot p \land \lnot q) \lor (q \land p) \equiv & &\text{By commutativity of $\land$} \\ & &\square \end{align} $$

I've assumed throughout $\land$ to have higher precedence than $\lor$.


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