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

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

I try to prove that $(-b<a) \land (a < b) \implies \vert a\vert<b$ 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 $a<b \implies \vert a \vert <b$.

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

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?

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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 $(P \land Q) \Rightarrow H$

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  • $\begingroup$ 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? $\endgroup$ – masiewpao Nov 17 '19 at 1:02
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    $\begingroup$ @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 $(P \land \neg P) \Rightarrow P$. $\endgroup$ – Bram28 Nov 17 '19 at 1:04
  • $\begingroup$ Ahh, OK that makes perfect sense thank you. $\endgroup$ – masiewpao Nov 17 '19 at 1:04
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    $\begingroup$ @masiewpao You're welcome! :) $\endgroup$ – Bram28 Nov 17 '19 at 1:06

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