Theorem hbra1 2664

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 2620	. 2 ⊢ (∀x ∈ A φ ↔ ∀x(x ∈ A → φ))
2		hba1 1786	. 2 ⊢ (∀x(x ∈ A → φ) → ∀x∀x(x ∈ A → φ))
3	1, 2	hbxfrbi 1568	1 ⊢ (∀x ∈ A φ → ∀x∀x ∈ A φ)

Syntax hints:   → wi 4  ∀wal 1540   ∈ wcel 1710  ∀wral 2615

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-11 1746

This theorem depends on definitions:  df-bi 177  df-ex 1542  df-ral 2620

This theorem is referenced by: (None)