Theorem 2ralbiim 3142

Description: Split a biconditional and distribute two restricted universal quantifiers, analogous to 2albiim 1923 and ralbiim 3125. (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 3125	. . 3 ⊢ (∀𝑦 ∈ 𝐵 (𝜑 ↔ 𝜓) ↔ (∀𝑦 ∈ 𝐵 (𝜑 → 𝜓) ∧ ∀𝑦 ∈ 𝐵 (𝜓 → 𝜑)))
2	1	ralbii 3109	. 2 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ↔ 𝜓) ↔ ∀𝑥 ∈ 𝐴 (∀𝑦 ∈ 𝐵 (𝜑 → 𝜓) ∧ ∀𝑦 ∈ 𝐵 (𝜓 → 𝜑)))
3		r19.26 3123	. 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 278	1 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ↔ 𝜓) ↔ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 → 𝜓) ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜓 → 𝜑)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wral 3077

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  df-ral 3078

This theorem is used by:  disjimeceqbi  39738  thincciso  50560