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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?

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@user18921, done with editing the question. – user11235813 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. – user11235813 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. – user11235813 Mar 15 '14 at 4:50
@JoseArnaldoDris, no worries, thanks for editing. – goblin Mar 15 '14 at 4:57
up vote 5 down vote accepted

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 $Q$.

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Thank you for your answer, @user18921 - this is just what I needed! =) – user11235813 Mar 15 '14 at 5:00
@JoseArnaldoDris, sweet. I love being useful. =) – goblin Mar 15 '14 at 5:01

Suppose that we examine rolling a 6 sided die.

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.

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Thanks @Vladhagen! – user11235813 Mar 15 '14 at 5:01

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