Answer: The given logical equivalence is proved below.
Step-by-step explanation: We are given to use truth tables to show the following logical equivalence :
P ⇔ Q ≡ (∼P ∨ Q)∧(∼Q ∨ P)
We know that
two compound propositions are said to be logically equivalent if they have same corresponding truth values in the truth table.
The truth table is as follows :
P Q ∼P ∼Q P⇔ Q ∼P ∨ Q ∼Q ∨ P (∼P ∨ Q)∧(∼Q ∨ P)
T T F F T T T T
T F F T F F T F
F T T F F T F F
F F T T T T T T
Since the corresponding truth vales for P ⇔ Q and (∼P ∨ Q)∧(∼Q ∨ P) are same, so the given propositions are logically equivalent.
Thus, P ⇔ Q ≡ (∼P ∨ Q)∧(∼Q ∨ P).
I think it’s 700, but I’m not positive
Look kid, this question has been up for 14 hours think its time to let it go......... hoped i helped. ✔verified✔
Slope is y2 - y1/ x2-x1 so # 1 would be 7-0/6-0 or 7/6
