Show that (∃x)[P (x) ∧ Q(x)] ⇒ (∃x)P (x) ∧ (∃x)Q(x)

asked by guest
on Nov 26, 2024 at 11:47 pm



Mathbot Says...

I wasn't able to parse your question, but the HE.NET team is hard at work making me smarter.


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