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15 December, 12:12

Identify the error in this argument that supposedly shows that if ∃xp (x) ∧ ∃xq (x) is true then ∃x (p (x) ∧ q (x)) is true

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  1. 15 December, 12:18
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    The problem is that ∃xp (x) is true, and ∃xq (x) is true means that for some x=x1, p (x) is true, and for some x=x2, q (x) is true. There is no guarantee that x1=x2, although it is not impossible.

    However, ∃x (p (x) ∧ q (x)) means that for the same x=x0, p (x0) is true, and q (x0) is true. This cannot be implied from the first proposition, therefore

    ∃xp (x) ∧ ∃xq (x) = > ∃x (p (x) ∧ q (x)) is a false statement.
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