Theorem rexbi 3119

Description: Distribute restricted quantification over a biconditional. (Contributed by Scott Fenton, 7-Aug-2024.) (Proof shortened by Wolf Lammen, 3-Nov-2024.)

Assertion

Ref	Expression
rexbi	⊢ (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓) → (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥 ∈ 𝐴 𝜓))

Proof of Theorem rexbi

Step	Hyp	Ref	Expression
1		biimp 218	. . . 4 ⊢ ((𝜑 ↔ 𝜓) → (𝜑 → 𝜓))
2	1	ralimi 3100	. . 3 ⊢ (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓) → ∀𝑥 ∈ 𝐴 (𝜑 → 𝜓))
3		rexim 3104	. . 3 ⊢ (∀𝑥 ∈ 𝐴 (𝜑 → 𝜓) → (∃𝑥 ∈ 𝐴 𝜑 → ∃𝑥 ∈ 𝐴 𝜓))
4	2, 3	syl 18	. 2 ⊢ (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓) → (∃𝑥 ∈ 𝐴 𝜑 → ∃𝑥 ∈ 𝐴 𝜓))
5		biimpr 223	. . . 4 ⊢ ((𝜑 ↔ 𝜓) → (𝜓 → 𝜑))
6	5	ralimi 3100	. . 3 ⊢ (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓) → ∀𝑥 ∈ 𝐴 (𝜓 → 𝜑))
7		rexim 3104	. . 3 ⊢ (∀𝑥 ∈ 𝐴 (𝜓 → 𝜑) → (∃𝑥 ∈ 𝐴 𝜓 → ∃𝑥 ∈ 𝐴 𝜑))
8	6, 7	syl 18	. 2 ⊢ (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓) → (∃𝑥 ∈ 𝐴 𝜓 → ∃𝑥 ∈ 𝐴 𝜑))
9	4, 8	impbid 215	1 ⊢ (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓) → (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥 ∈ 𝐴 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wral 3077  ∃wrex 3087

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-ex 1813  df-ral 3078  df-rex 3088

This theorem is used by:  ralrexbid  3120