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Metamath Proof Explorer

Theorem rexbi

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

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

Proof

Step Hyp Ref Expression
1 biimp ⊢ ( ( 𝜑 ↔ 𝜓 ) → ( 𝜑 → 𝜓 ) )
2 1 ralimi ⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 ↔ 𝜓 ) → ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝜓 ) )
3 rexim ⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝜓 ) → ( ∃ 𝑥 ∈ 𝐴 𝜑 → ∃ 𝑥 ∈ 𝐴 𝜓 ) )
4 2 3 syl ⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 ↔ 𝜓 ) → ( ∃ 𝑥 ∈ 𝐴 𝜑 → ∃ 𝑥 ∈ 𝐴 𝜓 ) )
5 biimpr ⊢ ( ( 𝜑 ↔ 𝜓 ) → ( 𝜓 → 𝜑 ) )
6 5 ralimi ⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 ↔ 𝜓 ) → ∀ 𝑥 ∈ 𝐴 ( 𝜓 → 𝜑 ) )
7 rexim ⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜓 → 𝜑 ) → ( ∃ 𝑥 ∈ 𝐴 𝜓 → ∃ 𝑥 ∈ 𝐴 𝜑 ) )
8 6 7 syl ⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 ↔ 𝜓 ) → ( ∃ 𝑥 ∈ 𝐴 𝜓 → ∃ 𝑥 ∈ 𝐴 𝜑 ) )
9 4 8 impbid ⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 ↔ 𝜓 ) → ( ∃ 𝑥 ∈ 𝐴 𝜑 ↔ ∃ 𝑥 ∈ 𝐴 𝜓 ) )