Theorem ralbiim 2685

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 392	. . 3 ⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)))
2	1	ralbii 2556	. 2 ⊢ (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓) ↔ ∀𝑥 ∈ 𝐴 ((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)))
3		r19.26 2677	. 2 ⊢ (∀𝑥 ∈ 𝐴 ((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) ↔ (∀𝑥 ∈ 𝐴 (𝜑 → 𝜓) ∧ ∀𝑥 ∈ 𝐴 (𝜓 → 𝜑)))
4	2, 3	bitri 184	1 ⊢ (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓) ↔ (∀𝑥 ∈ 𝐴 (𝜑 → 𝜓) ∧ ∀𝑥 ∈ 𝐴 (𝜓 → 𝜑)))

This proof depends on syntax axioms:   → wi 4   ∧ wa 104   ↔ wb 105  ∀wral 2528

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-gen 1502  ax-4 1563  ax-17 1579

This proof depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-ral 2533

This theorem is used by:  eqreu  3018