I tried to prove the statement that P ∧ (Q ∨ R) is logically equivalent to (P ∧ Q) ∨ (P ∧ R). I have just started my first proof class so I apologize if the proof is bad/is repetitive. Please see my proof below: Suppose that P ∧ (Q ∨ R) is true. Then P is true and Q ∨ R is true. Since Q ∨ R is true, at least one of Q or R is true.
Case 1: Suppose Q is true. Therefore P∧Q is true so (P ∧ Q) ∨ (P ∧ R) is true.
Case 2: Suppose R is true. Therefore P∧R is true so (P ∧ Q) ∨ (P ∧ R) is true.
Hence, (P ∧ Q) ∨ (P ∧ R) is true in either case. Thus if P ∧ (Q ∨ R) is true, then (P ∧ Q) ∨ (P ∧ R) is true.
Conversely, suppose that (P ∧ Q) ∨ (P ∧ R) is true. Thus at least one of the sentences P ∧ Q or P ∧ R is true.
Case 1: Suppose that P ∧ Q is true. Therefore P is true and Q is true so Q∨R is true. Since P is true and Q∨R is true, P ∧ (Q ∨ R) is true.
Case 2: Suppose that P ∧ R is true. Therefore P is true and R is true so Q∨R is true. Since P is true and Q∨R is true, P ∧ (Q ∨ R) is true.
Hence, P ∧ (Q ∨ R) is true in either case. Therefore it follows that P ∧ (Q ∨ R) is true exactly when the sentence (P ∧ Q) ∨ (P ∧ R) is true. By elimination, if P ∧ (Q ∨ R) is false, then (P ∧ Q) ∨ (P ∧ R) is false, and vice versa. Thus P ∧ (Q ∨ R) is logically equivalent to (P ∧ Q) ∨ (P ∧ R).
Yes. This is 0-1 enshorted version. Normally we should consider 8 cases concerned with logical values of P,Q and R and proving in both cases logical values of equivalent sentences are the same.
But we haven't to consider all cases. We can do that way like you did. We think just what we need to sentence to be true: they just to have both sides of equivalence have to the same value. So it's shrink to 4 cases