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How to combine an "exists" and a "forall" expressions for a same variable?

ex.:

$(\forall e \mid e \in S \Rightarrow e \in T) \vee(\exists e \mid e \in S \wedge e \notin T)$

Combining $2$ foralls or $2$ exists is simple, but what if it is a mix of both?

lolgab123
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1 Answers1

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Well, your example is a contradiction. Consequently, if you're trying to combine those two, in particular, then you should stop trying (or, rather, show that there is a contradiction).

More generally, you just need to determine under what circumstances both may hold, if any such circumstances exist. One thing that often helps is the equivalence (assuming excluded middle) of the following two statement types (for any statement $\Phi$ depending on $p\in J$):

  1. $\forall p\in J,\Phi(j)$
  2. $\neg\left(\exists p\in J:\Phi(j)\right)$

Added: As pointed out in the comments below, my first paragraph cannot be more wrong! Rather, the combined statement is simply true, regardless of $e, S$ and $T.$

Cameron Buie
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    It's not a contradiction; it's a tautology. Stated in English, it says: "Either $S$ is a subset of $T$, or it is not." – Wildcard Oct 03 '17 at 03:31
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    @Wildcard: You are quite right! I thought the connective was an "and," rather than an "or." /sigh/ – Cameron Buie Oct 03 '17 at 03:33