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Theorem 2albiim 1393
Description: Split a biconditional and distribute 2 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 1392 . . 3 (∀𝑦(𝜑𝜓) ↔ (∀𝑦(𝜑𝜓) ∧ ∀𝑦(𝜓𝜑)))
21albii 1375 . 2 (∀𝑥𝑦(𝜑𝜓) ↔ ∀𝑥(∀𝑦(𝜑𝜓) ∧ ∀𝑦(𝜓𝜑)))
3 19.26 1386 . 2 (∀𝑥(∀𝑦(𝜑𝜓) ∧ ∀𝑦(𝜓𝜑)) ↔ (∀𝑥𝑦(𝜑𝜓) ∧ ∀𝑥𝑦(𝜓𝜑)))
42, 3bitri 177 1 (∀𝑥𝑦(𝜑𝜓) ↔ (∀𝑥𝑦(𝜑𝜓) ∧ ∀𝑥𝑦(𝜓𝜑)))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 101  wb 102  wal 1257
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-5 1352  ax-gen 1354
This theorem depends on definitions:  df-bi 114
This theorem is referenced by:  sbnf2  1873  eqopab2b  4044  eqrel  4457  eqrelrel  4469  eqoprab2b  5591
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