Suppose you want to prove a theorem of the form p → (q ∨ r). prove that this is equivalent to showing that (p ∧ ¬q) → r.

p → (q ∨ r) ≡ ¬p ∨ (q ∨ r)implication rule

≡ (¬p ∨ q) ∨ rassociative rule

≡ ¬(p ∧ ¬q) ∨ rㅤㅤㅤ ㅤ how did i get this?

≡ (p ∧ ¬q) → rimplication rule

what rule was used to get ¬(p ∧ ¬q) ∨ r from (¬p ∨ q) ∨ r ?

You "distributed" the negation inside the parenthesis. Let's focus on the part.

We can distribute the negation inside the parenthesis by negating every term, and switching and with or, and vice versa. So, you have

Of course, writing is improper, I was just stressing the fact that we distributed the negation to every term in the parenthesis. So, since a double negation cancels out, you're left with

which of course implies that

Edit: I actually derived the expression in the opposite order, but since they are equivalent, it doesn't really matters where you start from and where you get to. Anyway, if you actually want to derive from , you simply have to negate the first parenthesis and distribute the negation as discussed.

x+y = 29, 5x + 2y = 100

step-by-step explanation:

i'm guessing it was a multiple first option was the correct answer