Theorem rexbi

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

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

Proof

Step	Hyp	Ref	Expression
1		ralbi	⊢ ( ∀ 𝑥 ∈ 𝐴 ( ¬ 𝜑 ↔ ¬ 𝜓 ) → ( ∀ 𝑥 ∈ 𝐴 ¬ 𝜑 ↔ ∀ 𝑥 ∈ 𝐴 ¬ 𝜓 ) )
2	1	notbid	⊢ ( ∀ 𝑥 ∈ 𝐴 ( ¬ 𝜑 ↔ ¬ 𝜓 ) → ( ¬ ∀ 𝑥 ∈ 𝐴 ¬ 𝜑 ↔ ¬ ∀ 𝑥 ∈ 𝐴 ¬ 𝜓 ) )
3		notbi	⊢ ( ( 𝜑 ↔ 𝜓 ) ↔ ( ¬ 𝜑 ↔ ¬ 𝜓 ) )
4	3	ralbii	⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 ↔ 𝜓 ) ↔ ∀ 𝑥 ∈ 𝐴 ( ¬ 𝜑 ↔ ¬ 𝜓 ) )
5		dfrex2	⊢ ( ∃ 𝑥 ∈ 𝐴 𝜑 ↔ ¬ ∀ 𝑥 ∈ 𝐴 ¬ 𝜑 )
6		dfrex2	⊢ ( ∃ 𝑥 ∈ 𝐴 𝜓 ↔ ¬ ∀ 𝑥 ∈ 𝐴 ¬ 𝜓 )
7	5 6	bibi12i	⊢ ( ( ∃ 𝑥 ∈ 𝐴 𝜑 ↔ ∃ 𝑥 ∈ 𝐴 𝜓 ) ↔ ( ¬ ∀ 𝑥 ∈ 𝐴 ¬ 𝜑 ↔ ¬ ∀ 𝑥 ∈ 𝐴 ¬ 𝜓 ) )
8	2 4 7	3imtr4i	⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 ↔ 𝜓 ) → ( ∃ 𝑥 ∈ 𝐴 𝜑 ↔ ∃ 𝑥 ∈ 𝐴 𝜓 ) )