Theorem 2albiim 1989

Description: Split a biconditional and distribute two quantifiers. (Contributed by NM, 3-Feb-2005.)

Assertion

Ref	Expression
2albiim	⊢ (∀𝑥∀𝑦(𝜑 ↔ 𝜓) ↔ (∀𝑥∀𝑦(𝜑 → 𝜓) ∧ ∀𝑥∀𝑦(𝜓 → 𝜑)))

Proof of Theorem 2albiim

Step	Hyp	Ref	Expression
1		albiim 1988	. . 3 ⊢ (∀𝑦(𝜑 ↔ 𝜓) ↔ (∀𝑦(𝜑 → 𝜓) ∧ ∀𝑦(𝜓 → 𝜑)))
2	1	albii 1915	. 2 ⊢ (∀𝑥∀𝑦(𝜑 ↔ 𝜓) ↔ ∀𝑥(∀𝑦(𝜑 → 𝜓) ∧ ∀𝑦(𝜓 → 𝜑)))
3		19.26 1969	. 2 ⊢ (∀𝑥(∀𝑦(𝜑 → 𝜓) ∧ ∀𝑦(𝜓 → 𝜑)) ↔ (∀𝑥∀𝑦(𝜑 → 𝜓) ∧ ∀𝑥∀𝑦(𝜓 → 𝜑)))
4	2, 3	bitri 267	1 ⊢ (∀𝑥∀𝑦(𝜑 ↔ 𝜓) ↔ (∀𝑥∀𝑦(𝜑 → 𝜓) ∧ ∀𝑥∀𝑦(𝜓 → 𝜑)))

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