There is this lemma: Let $\Sigma\subset \textrm{Prop}(A)$ and $p, q \in \textrm{Prop}(A)$. Then $\Sigma\models p \implies \Sigma\models p\vee q$. I can't figure out a counterexample for the opposite implication ($\textrm{Prop(A)}$ denotes the set of propositions and $A$ is a set of propositional atoms.

Thanks for help.



2 Answers 2


Let $q=\lnot p$.${}{}{}{}{}{}{}{}$

  • $\begingroup$ I think this is the smallest complete answer, I have seen in the site, Andre. However, that "Let" could be also ommitted. ;-) +1 $\endgroup$
    – Mikasa
    Apr 13, 2013 at 5:21
  • $\begingroup$ There have been shorter answers, a famous one by did, for example. About the suggestion to let go of let, I think one is not supposed to start a sentence with math symbols. $\endgroup$ Apr 13, 2013 at 5:25
  • $\begingroup$ Thanks. Yes. you are right. We cannot start such that in Maths at least. $\endgroup$
    – Mikasa
    Apr 13, 2013 at 5:27
  • $\begingroup$ @BabakS. There is math.stackexchange.com/a/74383/462 and math.stackexchange.com/a/180649/462 (and look at the comments on that one.) $\endgroup$ Apr 13, 2013 at 15:24

$${\bf \Sigma \models p \lor q \overset{?}\implies \Sigma \models p }\tag{${\bf converse}$}$$

What if $\;p\;$ is false and $\;q\;$ is true?: Suppose, e.g., $$\bf \text{ Suppose}\;\;\; q \;= \;\lnot p$$

$\quad$ Then your stated lemma: $\;\Sigma\models p \implies \Sigma\models p\vee q\;$ certainly holds.

But its converse (highlighted) certainly fails to hold.

  • $\begingroup$ Thank you very much, it's quite clear. $\endgroup$
    – pizet
    Apr 12, 2013 at 23:07
  • $\begingroup$ @amWhy: Yes, very clear amd it is nice to get that feedback from OP! +1 $\endgroup$
    – Amzoti
    Apr 13, 2013 at 2:29

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