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Theorem eximp-surprise2 46446
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 46445, which shows what implication really expands to. See also empty-surprise 46443. (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 864 . . 3 𝜑 → (¬ 𝜑𝜓))
31, 2eximii 1839 . 2 𝑥𝜑𝜓)
4 eximp-surprise 46445 . 2 (∃𝑥(𝜑𝜓) ↔ ∃𝑥𝜑𝜓))
53, 4mpbir 230 1 𝑥(𝜑𝜓)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wo 844  wex 1782
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812
This theorem depends on definitions:  df-bi 206  df-or 845  df-ex 1783
This theorem is referenced by: (None)
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