Vtyjkyuk

I'm trying to work out a class exercise and I've got myself stuck. Any help would be appreciated. Thank you.

I am to use use algebra of propositions to solve the following problem:

Show the below is true by the algebra of propositions:

(¬p ∨ q) ∧ (p ∨ ¬q) ≡ (p ∨ q) → (p ∧ q)

Here's what I have done so far:

Conditional and Biconditional: (¬p ∨ q) ∧ (p ∨ ¬q) ≡ (p → q) ∧ (q → p)

Current Conversion (No proof it is true yet): (p → q) ∧ (q → p) ≡ (p ∨ q) → (p ∧ q)

Conditional and Biconditional: (p → q) ∧ (q → p) ≡ p ↔ q

Current Conversion (No proof it is true yet): p ↔ q ≡ (p ∨ q) → (p ∧ q)

I have ran them through an online calculator for this type of thing and they are true but I cannot figure out the steps using the different Laws of Algebra of Propositions. If anyone could explain anything, I would be very grateful.

If you can use the following two equivalences:

$$p leftrightarrow q equiv (p rightarrow q) land (q rightarrow p)$$

$$p leftrightarrow q equiv (p land q) lor (neg p lor neg p)$$

Then you are indeed well on your way:

$$(neg p lor q) land (p lor neg q) text (Commutation)$$

$$(neg p lor q) land (neg q lor p) text (Implication)$$

$$(p rightarrow q) land (q rightarrow p) text (Equivalence)$$

$$p leftrightarrow q equiv text (Equivalence)$$

$$(p land q) lor (neg p land neg q) text (DeMorgan)$$

$$(p land q) lor neg (p lor q) text (Commutation)$$

$$neg (p land q) lor (p land q) text (Implication)$$

$$(p land q) to (p land q)$$

However, if you don't have both of those equivalences regarding the biconditional, then you'll have to do it the hard way, and starting at the left will actually be a little easier to follow:

$$(p lor q) to (p land q) text (Implication)$$

$$neg (p lor q) lor (p land q) equiv text (DeMorgan)$$

$$(neg p land neg q) lor (p land q) equiv text (Distribution)$$

$$((neg p land neg q) lor p) land ((neg p land neg q)lor q) equiv text (Distribution x 2)$$

$$((neg p lor p) land (neg q lor p)) land ((neg p lor q) land (neg q lor q)) equiv text (Commutation x 3)$$

$$((p lor neg p) land (p lor neg q)) land ((neg p lor q) land (q lor neg q)) equiv text (Complement x 2)$$

$$(top land (p lor neg q)) land ((neg p lor q) land top) equiv text (Identity x 2)$$

$$(p lor neg q) land (neg p lor q) equiv text (Commutation)$$

$$(neg p lor q) land (p lor neg q)$$