Theorem hbra1 2500

Description: x is not free in A. x e. A ph. (Contributed by NM, 18-Oct-1996.)

Assertion

Ref	Expression
hbra1	|- ( A. x e. A ph -> A. x A. x e. A ph )

Proof of Theorem hbra1

Step	Hyp	Ref	Expression
1		df-ral 2453	. 2 |- ( A. x e. A ph <-> A. x ( x e. A -> ph ) )
2		hba1 1533	. 2 |- ( A. x ( x e. A -> ph ) -> A. x A. x ( x e. A -> ph ) )
3	1, 2	hbxfrbi 1465	1 |- ( A. x e. A ph -> A. x A. x e. A ph )

Syntax hints:   -> wi 4   A.wal 1346   e. wcel 2141   A.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)