Recent content by the baby boy

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    I can't think of a counterexample to disprove this set theory theorem

    Could you further clarify how the union of a family set consisting of just {∅} becomes ∅? From my previous example, what would ∪F be? Family of sets F = {{1,2,3},{4,5,6},{∅}} ∪F = {1, 2, 3, 4, 5, 6} or {∅, 1, 2, 3, 4, 5, 6}?
  2. T

    I can't think of a counterexample to disprove this set theory theorem

    So empty sets are not counted in a union, but they are in an interception?
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    I can't think of a counterexample to disprove this set theory theorem

    Just so I understand, an empty set in a family of sets would be the following?: Family of sets F = {{1,2,3},{4,5,6},{∅}}? Suppose I included an empty set in both of the families, would the intersection still be a disjoint or would it be the set {∅}?
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    I can't think of a counterexample to disprove this set theory theorem

    I can't think of a counterexample to disprove this set theory "theorem" Assume F and G are families of sets. IF \cupF \bigcap \cupG = ∅ (disjoint), THEN F \bigcap G are disjoint as well.
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    Help with proving Biconditional equivalence

    Hello, Thank you all for helping me with this. I am only self-learning basic logic from a book, and the author never worked out distribution of all of the content in the parentheses to another, so I only thought you could distribute one letter at a time. I have a new problem: Prove (P...
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    Help with proving Biconditional equivalence

    How can I use the distributive law here? I mean, I don't see a common letter with a connective to factor out.
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    Help with proving Biconditional equivalence

    show that P \leftrightarrow Q is equal to (P\wedgeQ) \vee (\negP \wedge\negQ) (P→Q) \wedge (Q→P) (\negP\veeQ) \wedge (\negQ\veeP) [\neg(P\wedge\negQ)\wedge\neg(Q\wedge\negP)] \neg[(P\wedge\negQ)\vee(Q\wedge\negP)] I don't know which law to use from this point on to prove the...
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