Proving the equivalence without making use of Truth Tables How would I prove this without using the truth table? If anyone can help me with this it would be greatly appreciated
$$(¬A \lor B) \lor (¬B \lor ¬A) ≡ ¬A$$
This is what I got so far and I'm stuck after this
$$(A → B) \lor ¬(A \land B) ≡ ¬A$$
 A: The equivalence you have written does not hold.
Counterexample: $A$ and $B$ are both True. Then the left hand side evaluates to $(F \lor T) \lor (F \lor F)=T \lor F=T$, while the right hand side is $F$.
What does hold is that $(\neg A \lor B) \color{red}\land (\neg B \lor \neg A) \equiv \neg A$
Assuming you made a typo, and that that is what you needed to show, it
is easily shown as follows:
$$(\neg A \lor B) \land (\neg B \lor \neg A) \overset{Commutation}{\equiv}$$
$$(\neg A \lor B ) \land (\neg A \lor \neg B)\overset{Distribution}{\equiv}$$
$$\neg A \lor (B \land \neg B) \overset{Complement}{ \equiv}$$
$$\neg A \lor \bot \overset{Identity}{\equiv}$$
$$\neg A$$
A: Consider $A$- true and $B$ - true:
$$(\underbrace{(F)}_{\sim A}\vee\underbrace{(V)}_{B})\vee(\underbrace{(F)}_{\sim B}\vee\underbrace{(F)}_{\sim A}) \equiv (V)\vee(F) \equiv (V) \equiv A \not\equiv \sim A$$
A: The expression on the left-hand side is a tautology (which means it is always true). So, the logical equation you gave is not true.
Use associativity of $\lor$:
$$(¬A \lor B) \lor (¬B \lor ¬A) = ¬A \lor (B \lor ¬B) \lor ¬A = ¬A \lor T \lor ¬A = T$$
