Theorem rexbi 3154

Description: Distribute restricted quantification over a biconditional. (Contributed by Scott Fenton, 7-Aug-2024.)

Assertion

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

Proof of Theorem rexbi

Step	Hyp	Ref	Expression
1		ralbi 3080	. . 3 ⊢ (∀𝑥 ∈ 𝐴 (¬ 𝜑 ↔ ¬ 𝜓) → (∀𝑥 ∈ 𝐴 ¬ 𝜑 ↔ ∀𝑥 ∈ 𝐴 ¬ 𝜓))
2	1	notbid 321	. 2 ⊢ (∀𝑥 ∈ 𝐴 (¬ 𝜑 ↔ ¬ 𝜓) → (¬ ∀𝑥 ∈ 𝐴 ¬ 𝜑 ↔ ¬ ∀𝑥 ∈ 𝐴 ¬ 𝜓))
3		notbi 322	. . 3 ⊢ ((𝜑 ↔ 𝜓) ↔ (¬ 𝜑 ↔ ¬ 𝜓))
4	3	ralbii 3078	. 2 ⊢ (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓) ↔ ∀𝑥 ∈ 𝐴 (¬ 𝜑 ↔ ¬ 𝜓))
5		dfrex2 3151	. . 3 ⊢ (∃𝑥 ∈ 𝐴 𝜑 ↔ ¬ ∀𝑥 ∈ 𝐴 ¬ 𝜑)
6		dfrex2 3151	. . 3 ⊢ (∃𝑥 ∈ 𝐴 𝜓 ↔ ¬ ∀𝑥 ∈ 𝐴 ¬ 𝜓)
7	5, 6	bibi12i 343	. 2 ⊢ ((∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥 ∈ 𝐴 𝜓) ↔ (¬ ∀𝑥 ∈ 𝐴 ¬ 𝜑 ↔ ¬ ∀𝑥 ∈ 𝐴 ¬ 𝜓))
8	2, 4, 7	3imtr4i 295	1 ⊢ (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓) → (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥 ∈ 𝐴 𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209  ∀wral 3051  ∃wrex 3052

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817

This theorem depends on definitions:  df-bi 210  df-an 400  df-ex 1788  df-ral 3056  df-rex 3057

This theorem is referenced by: (None)