Theorem 2albiim 1417

Description: Split a biconditional and distribute 2 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 1416	. . 3 ⊢ (∀𝑦(𝜑 ↔ 𝜓) ↔ (∀𝑦(𝜑 → 𝜓) ∧ ∀𝑦(𝜓 → 𝜑)))
2	1	albii 1399	. 2 ⊢ (∀𝑥∀𝑦(𝜑 ↔ 𝜓) ↔ ∀𝑥(∀𝑦(𝜑 → 𝜓) ∧ ∀𝑦(𝜓 → 𝜑)))
3		19.26 1410	. 2 ⊢ (∀𝑥(∀𝑦(𝜑 → 𝜓) ∧ ∀𝑦(𝜓 → 𝜑)) ↔ (∀𝑥∀𝑦(𝜑 → 𝜓) ∧ ∀𝑥∀𝑦(𝜓 → 𝜑)))
4	2, 3	bitri 182	1 ⊢ (∀𝑥∀𝑦(𝜑 ↔ 𝜓) ↔ (∀𝑥∀𝑦(𝜑 → 𝜓) ∧ ∀𝑥∀𝑦(𝜓 → 𝜑)))

Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103  ∀wal 1282

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378

This theorem depends on definitions:  df-bi 115

This theorem is referenced by:  sbnf2  1898  eqopab2b  4034  eqrel  4447  eqrelrel  4459  eqoprab2b  5583