Determine if a formula is valid using equivalences
Examples:
Valid. Proof:
A → (B → A) ≡ ¬A ∨ (¬B ∨ A) ≡ ¬A ∨ A ∨ ¬B ≡ true ∨ ¬B ≡ true
A ∨ (A ∨ B). Answer: Not valid.
Model: A 7→ false, B 7→ false.
Here are my attempts at a couple of problems:
(1) (A ∧ B) ∨ (A ∧ ¬B)
(A ∧ B) ∨ (A ∧ ¬B): Answer Invalid
Model: A→False, B→True
(2) (¬(A ∨ B)) → ¬B:
This one is incomplete
(¬(A ∨ B)) → ¬B ≡ ((A ∧ B)) → B ≡
I'm not really sure how to do proofs by equivalence. I know deMorgan's laws are used here, but I'm struggling to understand how to apply it to actual problems.
\begin{align}(A ∧ B) ∨ (A ∧ ¬B)&\equiv A\land(B\lor \lnot B)\tag{1) Distributive Law} \\ \\ &\equiv A \land \text{true}\tag{2, LEM}\\ \\ &\equiv A\end{align}
(1) So If $A$ is false, (and $B$ is either true or false), the expression evaluates to false. On the other hand, if $A$ is true, the statement is true. So the truth value of the statement is "contingent".
*NOTE: "LEM" is short for the Law of the Excluded Middle. $$ $$ \begin{align}(\lnot(A \lor B)) \to \lnot B &\equiv \lnot\lnot (A\lor B) \lor \lnot B \tag{2}\\ \\ &\equiv (A\lor B)\lor \lnot B\\ \\ &\equiv A \lor (B \lor \lnot B)\\ \\ &\equiv A \lor \text{true} \\ \\ &\equiv \text{true}\end{align}
Note that this statement (2), is a tautology, true no matter what the truth values assigned to $A, B$. I'll let you