Theorem r2allem 3165

Description: Lemma factoring out common proof steps of r2alf 3186 and r2al 3166. Introduced to reduce dependencies on axioms. (Contributed by Wolf Lammen, 9-Jan-2020.)

Hypothesis

Ref	Expression
r2allem.1	⊢ (∀𝑦(𝑥 ∈ 𝐴 → (𝑦 ∈ 𝐵 → 𝜑)) ↔ (𝑥 ∈ 𝐴 → ∀𝑦(𝑦 ∈ 𝐵 → 𝜑)))

Assertion

Ref	Expression
r2allem	⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 𝜑 ↔ ∀𝑥∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → 𝜑))

Proof of Theorem r2allem

Step	Hyp	Ref	Expression
1		df-ral 3111	. 2 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 𝜑 ↔ ∀𝑥(𝑥 ∈ 𝐴 → ∀𝑦 ∈ 𝐵 𝜑))
2		r2allem.1	. . . 4 ⊢ (∀𝑦(𝑥 ∈ 𝐴 → (𝑦 ∈ 𝐵 → 𝜑)) ↔ (𝑥 ∈ 𝐴 → ∀𝑦(𝑦 ∈ 𝐵 → 𝜑)))
3		impexp 454	. . . . 5 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → 𝜑) ↔ (𝑥 ∈ 𝐴 → (𝑦 ∈ 𝐵 → 𝜑)))
4	3	albii 1821	. . . 4 ⊢ (∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → 𝜑) ↔ ∀𝑦(𝑥 ∈ 𝐴 → (𝑦 ∈ 𝐵 → 𝜑)))
5		df-ral 3111	. . . . 5 ⊢ (∀𝑦 ∈ 𝐵 𝜑 ↔ ∀𝑦(𝑦 ∈ 𝐵 → 𝜑))
6	5	imbi2i 339	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → ∀𝑦 ∈ 𝐵 𝜑) ↔ (𝑥 ∈ 𝐴 → ∀𝑦(𝑦 ∈ 𝐵 → 𝜑)))
7	2, 4, 6	3bitr4i 306	. . 3 ⊢ (∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → 𝜑) ↔ (𝑥 ∈ 𝐴 → ∀𝑦 ∈ 𝐵 𝜑))
8	7	albii 1821	. 2 ⊢ (∀𝑥∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → 𝜑) ↔ ∀𝑥(𝑥 ∈ 𝐴 → ∀𝑦 ∈ 𝐵 𝜑))
9	1, 8	bitr4i 281	1 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 𝜑 ↔ ∀𝑥∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → 𝜑))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 399  ∀wal 1536   ∈ wcel 2111  ∀wral 3106

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811

This theorem depends on definitions:  df-bi 210  df-an 400  df-ral 3111

This theorem is referenced by:  r2al  3166  r2alf  3186