Theorem 2albiim 1923

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 1922	. . 3 ⊢ (∀𝑦(𝜑 ↔ 𝜓) ↔ (∀𝑦(𝜑 → 𝜓) ∧ ∀𝑦(𝜓 → 𝜑)))
2	1	albii 1852	. 2 ⊢ (∀𝑥∀𝑦(𝜑 ↔ 𝜓) ↔ ∀𝑥(∀𝑦(𝜑 → 𝜓) ∧ ∀𝑦(𝜓 → 𝜑)))
3		19.26 1903	. 2 ⊢ (∀𝑥(∀𝑦(𝜑 → 𝜓) ∧ ∀𝑦(𝜓 → 𝜑)) ↔ (∀𝑥∀𝑦(𝜑 → 𝜓) ∧ ∀𝑥∀𝑦(𝜓 → 𝜑)))
4	2, 3	bitri 278	1 ⊢ (∀𝑥∀𝑦(𝜑 ↔ 𝜓) ↔ (∀𝑥∀𝑦(𝜑 → 𝜓) ∧ ∀𝑥∀𝑦(𝜓 → 𝜑)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842

This proof depends on definitions:  df-bi 210  df-an 402

This theorem is used by:  sbnf2  2389  2eu6  2683  eqopab2bw  5531  eqopab2b  5535  eqrel  5768  eqrelrel  5781  eqoprab2bw  7487  eqoprab2b  7488  eqrelrd2  33097  eqrel2  39061  relcnveq2  39085  elrelscnveq2  39385  pm14.123a  45257