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Theorem 2albiim 1989
Description: Split a biconditional and distribute two quantifiers. (Contributed by NM, 3-Feb-2005.)
Assertion
Ref Expression
2albiim (∀𝑥𝑦(𝜑𝜓) ↔ (∀𝑥𝑦(𝜑𝜓) ∧ ∀𝑥𝑦(𝜓𝜑)))

Proof of Theorem 2albiim
StepHypRef Expression
1 albiim 1988 . . 3 (∀𝑦(𝜑𝜓) ↔ (∀𝑦(𝜑𝜓) ∧ ∀𝑦(𝜓𝜑)))
21albii 1915 . 2 (∀𝑥𝑦(𝜑𝜓) ↔ ∀𝑥(∀𝑦(𝜑𝜓) ∧ ∀𝑦(𝜓𝜑)))
3 19.26 1969 . 2 (∀𝑥(∀𝑦(𝜑𝜓) ∧ ∀𝑦(𝜓𝜑)) ↔ (∀𝑥𝑦(𝜑𝜓) ∧ ∀𝑥𝑦(𝜓𝜑)))
42, 3bitri 267 1 (∀𝑥𝑦(𝜑𝜓) ↔ (∀𝑥𝑦(𝜑𝜓) ∧ ∀𝑥𝑦(𝜓𝜑)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 198  wa 385  wal 1651
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905
This theorem depends on definitions:  df-bi 199  df-an 386
This theorem is referenced by:  sbnf2  2557  2eu6  2712  eqopab2b  5199  eqrel  5411  eqrelrel  5423  eqoprab2b  6945  eqrelrd2  29937  eqrel2  34557  relcnveq2  34580  elrelscnveq2  34729  pm14.123a  39395
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