Theorem hbra1 2500

Description: 𝑥 is not free in ∀𝑥 ∈ 𝐴𝜑. (Contributed by NM, 18-Oct-1996.)

Assertion

Ref	Expression
hbra1	⊢ (∀𝑥 ∈ 𝐴 𝜑 → ∀𝑥∀𝑥 ∈ 𝐴 𝜑)

Proof of Theorem hbra1

Step	Hyp	Ref	Expression
1		df-ral 2453	. 2 ⊢ (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))
2		hba1 1533	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝜑) → ∀𝑥∀𝑥(𝑥 ∈ 𝐴 → 𝜑))
3	1, 2	hbxfrbi 1465	1 ⊢ (∀𝑥 ∈ 𝐴 𝜑 → ∀𝑥∀𝑥 ∈ 𝐴 𝜑)

Syntax hints:   → wi 4  ∀wal 1346   ∈ wcel 2141  ∀wral 2448

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1440  ax-gen 1442  ax-ial 1527

This theorem depends on definitions:  df-bi 116  df-ral 2453

This theorem is referenced by: (None)