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Theorem eximp-surprise2 44253
 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(𝜑 → 𝜓) 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 44252, which shows what implication really expands to. See also empty-surprise 44250. (Contributed by David A. Wheeler, 18-Oct-2018.)
Hypothesis
Ref Expression
eximp-surprise2.1 𝑥 ¬ 𝜑
Assertion
Ref Expression
eximp-surprise2 𝑥(𝜑𝜓)

Proof of Theorem eximp-surprise2
StepHypRef Expression
1 eximp-surprise2.1 . . 3 𝑥 ¬ 𝜑
2 orc 853 . . 3 𝜑 → (¬ 𝜑𝜓))
31, 2eximii 1799 . 2 𝑥𝜑𝜓)
4 eximp-surprise 44252 . 2 (∃𝑥(𝜑𝜓) ↔ ∃𝑥𝜑𝜓))
53, 4mpbir 223 1 𝑥(𝜑𝜓)
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ∨ wo 833  ∃wex 1742 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1758  ax-4 1772 This theorem depends on definitions:  df-bi 199  df-or 834  df-ex 1743 This theorem is referenced by: (None)
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