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Theorem rexbi 3169
Description: Distribute restricted quantification over a biconditional. (Contributed by Scott Fenton, 7-Aug-2024.) (Proof shortened by Wolf Lammen, 3-Nov-2024.)
Assertion
Ref Expression
rexbi (∀𝑥𝐴 (𝜑𝜓) → (∃𝑥𝐴 𝜑 ↔ ∃𝑥𝐴 𝜓))

Proof of Theorem rexbi
StepHypRef Expression
1 biimp 214 . . . 4 ((𝜑𝜓) → (𝜑𝜓))
21ralimi 3086 . . 3 (∀𝑥𝐴 (𝜑𝜓) → ∀𝑥𝐴 (𝜑𝜓))
3 rexim 3168 . . 3 (∀𝑥𝐴 (𝜑𝜓) → (∃𝑥𝐴 𝜑 → ∃𝑥𝐴 𝜓))
42, 3syl 17 . 2 (∀𝑥𝐴 (𝜑𝜓) → (∃𝑥𝐴 𝜑 → ∃𝑥𝐴 𝜓))
5 biimpr 219 . . . 4 ((𝜑𝜓) → (𝜓𝜑))
65ralimi 3086 . . 3 (∀𝑥𝐴 (𝜑𝜓) → ∀𝑥𝐴 (𝜓𝜑))
7 rexim 3168 . . 3 (∀𝑥𝐴 (𝜓𝜑) → (∃𝑥𝐴 𝜓 → ∃𝑥𝐴 𝜑))
86, 7syl 17 . 2 (∀𝑥𝐴 (𝜑𝜓) → (∃𝑥𝐴 𝜓 → ∃𝑥𝐴 𝜑))
94, 8impbid 211 1 (∀𝑥𝐴 (𝜑𝜓) → (∃𝑥𝐴 𝜑 ↔ ∃𝑥𝐴 𝜓))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205  wral 3063  wrex 3064
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813
This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1784  df-ral 3068  df-rex 3069
This theorem is referenced by:  ralrexbid  3250
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