Answer:
(X and Y are real numbers)
p) X > 0
q) Y > 0
r) XY > 0
s) (X > 0 ∧ Y > 0) ∨ (X = Y = 0)
Step-by-step explanation:
Lets assume that a pair of the propositions are true, and we will show that the other 2 are also true. There are 6 possible cases:
1) If p and q are true, then XY has to be positive because it is the product of positive numbers. And the first option of proposition s is true
2) If p and r are true, then Y = XY/X > 0 because it is the division of tow positive numbers, thus q is true. Since p and q are true, then so it is s.
3) If p and s are true, then option 2 of S is impossible, thus X > 0 and Y>0, hence q is true. SInce p and q are true, so is r.
4) If q and r are true, we can obtain X by dividing the positive numbers XY and Y, thus X>0, and p and s are true.
5) is q and s are true, an argument similar at the one made in 3) shows that both p and r are true.
6) if r and s are true, then the second part of s cant be true, therefore X>0 and Y>0, as a consecuence, p and q are true.
Now, lets show that, individually, no proposition implies the others:
p clearly doesnt imply q. X can be positive while Y is negative. Similarly Y can be positive while X is negative, thus q doesnt imply p either.
If XY > 0, then it can be that both X and Y are negative, thus r might not imply neither p nor q.
and if only s is true, then it may be that the second part is the one that is true, thereofre X=Y=0 and p, q and r are all false in this case.
As a consecuence, you need 2 of the propositions to be true so that all 4 are true, but as we show above, by having any pair of the propositions true, then all 4 are automaticallyy true.
