Theorem 2albiim 1890

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 1889	. . 3 ⊢ (∀𝑦(𝜑 ↔ 𝜓) ↔ (∀𝑦(𝜑 → 𝜓) ∧ ∀𝑦(𝜓 → 𝜑)))
2	1	albii 1819	. 2 ⊢ (∀𝑥∀𝑦(𝜑 ↔ 𝜓) ↔ ∀𝑥(∀𝑦(𝜑 → 𝜓) ∧ ∀𝑦(𝜓 → 𝜑)))
3		19.26 1870	. 2 ⊢ (∀𝑥(∀𝑦(𝜑 → 𝜓) ∧ ∀𝑦(𝜓 → 𝜑)) ↔ (∀𝑥∀𝑦(𝜑 → 𝜓) ∧ ∀𝑥∀𝑦(𝜓 → 𝜑)))
4	2, 3	bitri 275	1 ⊢ (∀𝑥∀𝑦(𝜑 ↔ 𝜓) ↔ (∀𝑥∀𝑦(𝜑 → 𝜓) ∧ ∀𝑥∀𝑦(𝜓 → 𝜑)))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395  ∀wal 1538

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809

This theorem depends on definitions:  df-bi 207  df-an 396

This theorem is referenced by:  sbnf2  2357  2eu6  2651  eqopab2bw  5511  eqopab2b  5515  eqrel  5750  eqrelrel  5763  eqoprab2bw  7462  eqoprab2b  7463  eqrelrd2  32551  eqrel2  38294  relcnveq2  38318  elrelscnveq2  38491  pm14.123a  44421