# A question on logic

Suppose I know that the following implications are true:

$$P_1 \Longrightarrow (A \land B)$$ $$P_2 \Longrightarrow (A \land B)$$

for some premises $P_1, P_2$ and some conditions $A, B$.

Does it follow that $A \land B$ is true, if $P_1$ and $P_2$ are mutually exclusive conditions?

• @user18921, done with editing the question. – Jose Arnaldo Bebita-Dris Mar 15 '14 at 4:46
• Essentially, I am doing a brute-force analysis of all possible cases for a particular math problem. It all goes down to breaking the cases to the subcases $k = 1$ and $k > 1$. You can check out my preprints in the arXiv if you'd like to browse through my work so far. – Jose Arnaldo Bebita-Dris Mar 15 '14 at 4:47
• use a truth table? This looks like $P_1$ implies $A \land B$ but I don't know if $P_1$ and $P_2$ have their own truth table values? – usukidoll Mar 15 '14 at 4:48
• Thank you for the upvotes. I apologize for not being clear with phrasing my initial question earlier. – Jose Arnaldo Bebita-Dris Mar 15 '14 at 4:50
• @JoseArnaldoDris, no worries, thanks for editing. – goblin Mar 15 '14 at 4:57

No; given that $P_1$ and $P_2$ are mutually exclusive conditions, it is still possible that they're both false, and so we cannot deduce anything from $P_1\rightarrow Q,$ nor from $P_2 \rightarrow Q$. Explicitly (note that false implies false), the following is a counterexample to the conjecture:
$$P_1 = \mathrm{False}, \;P_2 = \mathrm{False}, \;Q = \mathrm{False}$$
Now on the other hand, if $P_1$ and $P_2$ are mutually exhaustive conditions (i.e., at least one of them is true), and if we know that $P_1 \rightarrow Q$ and $P_2 \rightarrow Q,$ then we may deduce $Q$.
Let $P_1$ be getting a 1 and let $P_2$ be rolling a 2. These are mutually exclusive events. Say that rolling a 1 or 2 guarantees winning \$20 (A) and getting a new haircut (B). But any other roll does not guarantee these things. So$A$and$B\$ might not necessarily always hold.