# What does it mean to "show algebraically" in propositional logic?

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?

• "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. Aug 14 '20 at 15:15

\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 $$\lor$$.