# How to use direct proof when the antecedent of an implication contains an 'and' statement?

In trying to prove the proposition $$\forall a \in \mathbb{R}, \forall b \in \mathbb{R}^+, \vert a\vert

I can do the $$\implies$$ direction, but am not sure if my reasoning for the other direction is correct.

I try to prove that $$(-b by cases. My question pertains specifically to proving the case in which $$a$$ is positive.

From writing out the truth tables I see that $$(P\land Q) \implies H$$ is logically equivalent to $$(P\implies H) \lor\neg Q$$, I just show that in the case $$a$$ is positive, then $$a = \vert a \vert$$, so $$a.

This seems to prove that $$P \implies H$$ is true, where $$P$$: $$a and $$H$$: $$\vert a \vert .

What am I wondering is if this is valid? It feels like I didn't use $$Q$$ in the reasoning to prove $$H$$, other than assuming it's true. By that I mean, in the proof I write out, "assume $$P$$ and $$Q$$...", but then in the reasoning I 'prove' the result by showing $$P$$ implies $$H$$ by itself.

Is this correct? I don't quite understand intuitively why it is valid to not 'do' anything with my assumption that $$Q$$ is true. Generally in direct proofs when we assume $$P$$ is true in order to derive $$Q$$, we use the information about $$P$$ to get there, and in this case it seems like I haven't done that.

I think my ultimate question is what is the general approach to solving statements of the logical form $$(P\land Q) \implies H$$?

Is it sufficient to prove $$P\implies H$$, since it is part of a compound 'or' statement?

Correct! If you can show that $$P \Rightarrow H$$, then it is also true that $$(P \land Q) \Rightarrow H$$, for if $$P \land Q$$ is true, then certainly $$P$$ is true, and so combined with $$P \Rightarrow H$$ you thus get $$H$$, as desired.

Or, as a more formal proof:

$$\begin{array}{lll} 1& P \Rightarrow H & \text{already shown}\\ 2& P \land Q & Assumption\\ 3& P & \land \ Elim 2\\ 4& H & MP \ 1,3\\ 5& (P \land Q) \Rightarrow H & \Rightarrow \ Intro 2-4\\ \end{array}$$

Or, more simply:

$$(P \land Q) \Rightarrow H$$ is equivalent to $$\neg (P \land Q) \lor H$$, and thus to $$\neg P \lor \neg Q \lor H$$. So, if you can show $$P \Rightarrow H$$, which is equivalent to $$\neg P \lor H$$, then it is certainly true that $$\neg P \lor \neg Q \lor H$$, and hence $$(P \land Q) \Rightarrow H$$

• Thanks very much! I wasn't sure if the reasoning was correct or not. If I could clarify one additional thing, is there not a risk that by not using all the conjuncts, then we could end up with a contradiction? For example we prove $P\implies Q$, but then since we don't use the other conjunct, we miss that $Q\implies \neg P$? Is this a valid concern? – masiewpao Nov 17 '19 at 1:02
• @masiewpao No need to be concerned about that. A 'problem' with the givens is not the same as a 'problem' with the argument. In fact, if there is a contradiction in the givens, then anything validly follows. Example: show that $(P \land \neg P) \Rightarrow P$. Well, since $P \Rightarrow P$, it should also be true that $(P \land \neg P) \Rightarrow P$. Does the fact that $P \land \neg P$ take anything away from that? No, because $(P \land \neg P) \Rightarrow \bot$, and $\bot \Rightarrow P$, and so $(P \land \neg P) \Rightarrow P$. – Bram28 Nov 17 '19 at 1:04
• Ahh, OK that makes perfect sense thank you. – masiewpao Nov 17 '19 at 1:04
• @masiewpao You're welcome! :) – Bram28 Nov 17 '19 at 1:06