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Theorem eximal 1698
 Description: A utility theorem. An interesting case is when the same formula is substituted for both 𝜑 and 𝜓, since then both implications express a type of non-freeness. See also alimex 1748. (Contributed by BJ, 12-May-2019.)
Assertion
Ref Expression
eximal ((∃𝑥𝜑𝜓) ↔ (¬ 𝜓 → ∀𝑥 ¬ 𝜑))

Proof of Theorem eximal
StepHypRef Expression
1 df-ex 1696 . . 3 (∃𝑥𝜑 ↔ ¬ ∀𝑥 ¬ 𝜑)
21imbi1i 338 . 2 ((∃𝑥𝜑𝜓) ↔ (¬ ∀𝑥 ¬ 𝜑𝜓))
3 con1b 347 . 2 ((¬ ∀𝑥 ¬ 𝜑𝜓) ↔ (¬ 𝜓 → ∀𝑥 ¬ 𝜑))
42, 3bitri 263 1 ((∃𝑥𝜑𝜓) ↔ (¬ 𝜓 → ∀𝑥 ¬ 𝜑))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 195  ∀wal 1473  ∃wex 1695 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8 This theorem depends on definitions:  df-bi 196  df-ex 1696 This theorem is referenced by:  ax5e  1829  19.23t  2066  19.23tOLD  2206  xfree2  28688  bj-nalnalimiOLD  31799  bj-exalimi  31801
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