I'm having a problem understanding something.

In an excercise the following statement is wrong and I'm having trouble understanding why.

"If $a$ and $b$ are propositions then $(a \lor b)$ is valid if $a$ is valid or $b$ is valid."

Why is it wrong? Is there some interpretation where this doesn't hold?

Sorry if this is trivial, but I'm new to logic and I am trying to understand things

  • 1
    $\begingroup$ The statement that you quote is not wrong. It would be wrong if you changed it to "$a \lor b$ is valid iff $a$ is valid or $b$ is valid". Have you copied it correctly? $\endgroup$
    – Rob Arthan
    Sep 8, 2016 at 19:53
  • $\begingroup$ Oh, yes you are right. And I think I can see why it's wrong that way. If $a \lor b$ is valid that means that one interpretation can verify $a$ and not $b$ and another can verify $b$ and not $a$. So that means that there is no need that either one be valid for all interpretations. Right? $\endgroup$ Sep 8, 2016 at 20:02
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    $\begingroup$ That's right. Over the integers $x \le 0 \lor 0 < x$ is valid but neither $x \le 0$ nor $0 < x$ is valid. $\endgroup$
    – Rob Arthan
    Sep 8, 2016 at 20:09
  • $\begingroup$ Thank you. I will accept it as an answer if you post it. Otherwise I guess I'll delete the question $\endgroup$ Sep 8, 2016 at 20:15
  • $\begingroup$ And thank you. I've expanded the comment into an answer. $\endgroup$
    – Rob Arthan
    Sep 8, 2016 at 20:27

1 Answer 1


It is true that $a \lor b$ is valid (i.e., holds in all interpretations) if either $a$ is valid or $b$ is valid. So, for example, $x^2 \ge 0 \lor x ^2 < 0$ is valid over the integers because $x^2 \ge 0$ is. The converse is not true: $a \lor b$ may be valid when neither $a$ nor $b$ on its own is valid, because $b$ may hold in the interpretations where $a$ does not. E.g., over the integers $x \le 0 \lor x > 0$ is valid, while neither $x\le 0$ nor $x > 0$ is valid.


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