Theorem ralbiim 2600

Description: Split a biconditional and distribute quantifier. (Contributed by NM, 3-Jun-2012.)

Assertion

Ref	Expression
ralbiim	⊢ (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓) ↔ (∀𝑥 ∈ 𝐴 (𝜑 → 𝜓) ∧ ∀𝑥 ∈ 𝐴 (𝜓 → 𝜑)))

Proof of Theorem ralbiim

Step	Hyp	Ref	Expression
1		dfbi2 386	. . 3 ⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)))
2	1	ralbii 2472	. 2 ⊢ (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓) ↔ ∀𝑥 ∈ 𝐴 ((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)))
3		r19.26 2592	. 2 ⊢ (∀𝑥 ∈ 𝐴 ((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) ↔ (∀𝑥 ∈ 𝐴 (𝜑 → 𝜓) ∧ ∀𝑥 ∈ 𝐴 (𝜓 → 𝜑)))
4	2, 3	bitri 183	1 ⊢ (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓) ↔ (∀𝑥 ∈ 𝐴 (𝜑 → 𝜓) ∧ ∀𝑥 ∈ 𝐴 (𝜓 → 𝜑)))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104  ∀wral 2444

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1435  ax-gen 1437  ax-4 1498  ax-17 1514

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-ral 2449

This theorem is referenced by:  eqreu  2918