Theorem 2albiim 1612

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 1611	. . 3 ⊢ (∀y(φ ↔ ψ) ↔ (∀y(φ → ψ) ∧ ∀y(ψ → φ)))
2	1	albii 1566	. 2 ⊢ (∀x∀y(φ ↔ ψ) ↔ ∀x(∀y(φ → ψ) ∧ ∀y(ψ → φ)))
3		19.26 1593	. 2 ⊢ (∀x(∀y(φ → ψ) ∧ ∀y(ψ → φ)) ↔ (∀x∀y(φ → ψ) ∧ ∀x∀y(ψ → φ)))
4	2, 3	bitri 240	1 ⊢ (∀x∀y(φ ↔ ψ) ↔ (∀x∀y(φ → ψ) ∧ ∀x∀y(ψ → φ)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557

This proof depends on definitions:  df-bi 177  df-an 360

This theorem is used by:  sbnf2  2108  2eu6  2289  eqrel  4846  eqopr  4848