Theorem 2ralbiim 3100

Description: Split a biconditional and distribute two restricted universal quantifiers, analogous to 2albiim 1893 and ralbiim 3099. (Contributed by Alexander van der Vekens, 2-Jul-2017.)

Assertion

Ref	Expression
2ralbiim	⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ↔ 𝜓) ↔ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 → 𝜓) ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜓 → 𝜑)))

Proof of Theorem 2ralbiim

Step	Hyp	Ref	Expression
1		ralbiim 3099	. . 3 ⊢ (∀𝑦 ∈ 𝐵 (𝜑 ↔ 𝜓) ↔ (∀𝑦 ∈ 𝐵 (𝜑 → 𝜓) ∧ ∀𝑦 ∈ 𝐵 (𝜓 → 𝜑)))
2	1	ralbii 3092	. 2 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ↔ 𝜓) ↔ ∀𝑥 ∈ 𝐴 (∀𝑦 ∈ 𝐵 (𝜑 → 𝜓) ∧ ∀𝑦 ∈ 𝐵 (𝜓 → 𝜑)))
3		r19.26 3095	. 2 ⊢ (∀𝑥 ∈ 𝐴 (∀𝑦 ∈ 𝐵 (𝜑 → 𝜓) ∧ ∀𝑦 ∈ 𝐵 (𝜓 → 𝜑)) ↔ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 → 𝜓) ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜓 → 𝜑)))
4	2, 3	bitri 274	1 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ↔ 𝜓) ↔ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 → 𝜓) ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜓 → 𝜑)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 396  ∀wral 3064

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812

This theorem depends on definitions:  df-bi 206  df-an 397  df-ral 3069

This theorem is referenced by:  thincciso  46330