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5:32 AM
12 hours ago, by user21820
If ¬∀x∈S (P(x)):
    If ¬∃x∈S (¬P(x)):
        ¬∃y∈S (¬P(y)). // Rename
        Given x∈S:
            If ¬P(x):
                ∃y∈S (¬P(y)). // Exists-Intro since x is unused.
                ¬∃y∈S (¬P(y)). // Restate since y does not occur in it.
            P(x).
        ∀x∈S (P(x)).
        ¬∀x∈S (P(x)). // Restate
        Contradiction.
    ∃x∈S (¬P(x)).
¬∀x∈S (P(x)) → ∃x∈S (¬P(x)).
Gah... I realize my comments in the above proof were wrong...
2 messages moved from Logic
 
 
11 hours later…
Anonymous
4:55 PM
7 messages moved to Trashcan
 

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