Theorem 2albiim 1966

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 1965	. . 3 ⊢ (∀𝑦(𝜑 ↔ 𝜓) ↔ (∀𝑦(𝜑 → 𝜓) ∧ ∀𝑦(𝜓 → 𝜑)))
2	1	albii 1896	. 2 ⊢ (∀𝑥∀𝑦(𝜑 ↔ 𝜓) ↔ ∀𝑥(∀𝑦(𝜑 → 𝜓) ∧ ∀𝑦(𝜓 → 𝜑)))
3		19.26 1947	. 2 ⊢ (∀𝑥(∀𝑦(𝜑 → 𝜓) ∧ ∀𝑦(𝜓 → 𝜑)) ↔ (∀𝑥∀𝑦(𝜑 → 𝜓) ∧ ∀𝑥∀𝑦(𝜓 → 𝜑)))
4	2, 3	bitri 264	1 ⊢ (∀𝑥∀𝑦(𝜑 ↔ 𝜓) ↔ (∀𝑥∀𝑦(𝜑 → 𝜓) ∧ ∀𝑥∀𝑦(𝜓 → 𝜑)))

Syntax hints:   → wi 4   ↔ wb 196   ∧ wa 383  ∀wal 1630

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886

This theorem depends on definitions:  df-bi 197  df-an 385

This theorem is referenced by:  sbnf2  2576  2eu6  2696  eqopab2b  5155  eqrel  5366  eqrelrel  5378  eqoprab2b  6878  eqrelrd2  29735  eqrel2  34392  relcnveq2  34418  elrelscnveq2  34566  pm14.123a  39128