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Theorem eximp-surprise2 41864
Description: Show that "there exists" with an implication is always true if there exists a situation where the antecedent is false.

Those inexperienced with formal notations of classical logic may use expressions combining "there exists" with implication. This is usually a mistake, because that combination does not mean what an inexperienced person might think it means. For example, if there is some object that does not meet the precondition 𝜑, then the expression 𝑥(𝜑𝜓) as a whole is always true, no matter what 𝜓 is (𝜓 could even be false, ). New users of formal notation who use "there exists" with an implication should consider if they meant "and" instead of "implies". See eximp-surprise 41863, which shows what implication really expands to. See also empty-surprise 41861. (Contributed by David A. Wheeler, 18-Oct-2018.)

Ref Expression
eximp-surprise2.1 𝑥 ¬ 𝜑
Ref Expression
eximp-surprise2 𝑥(𝜑𝜓)

Proof of Theorem eximp-surprise2
StepHypRef Expression
1 eximp-surprise2.1 . . 3 𝑥 ¬ 𝜑
2 orc 400 . . 3 𝜑 → (¬ 𝜑𝜓))
31, 2eximii 1761 . 2 𝑥𝜑𝜓)
4 eximp-surprise 41863 . 2 (∃𝑥(𝜑𝜓) ↔ ∃𝑥𝜑𝜓))
53, 4mpbir 221 1 𝑥(𝜑𝜓)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wo 383  wex 1701
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734
This theorem depends on definitions:  df-bi 197  df-or 385  df-ex 1702
This theorem is referenced by: (None)
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