Theorem hbre1 1692

Description: x is not free in E.x e. Aph.

Assertion

Ref	Expression
hbre1	|- (E.x e. A ph -> A.xE.x e. A ph)

Proof of Theorem hbre1

Step	Hyp	Ref	Expression
1		hbe1 1018	. 2 |- (E.x(x e. A /\ ph) -> A.xE.x(x e. A /\ ph))
2		df-rex 1653	. 2 |- (E.x e. A ph <-> E.x(x e. A /\ ph))
3	2	albii 1001	. 2 |- (A.xE.x e. A ph <-> A.xE.x(x e. A /\ ph))
4	1, 2, 3	3imtr4 219	1 |- (E.x e. A ph -> A.xE.x e. A ph)

Syntax hints:   -> wi 3   /\ wa 223  A.wal 956   e. wcel 960  E.wex 982  E.wrex 1649

This theorem is referenced by:  uniiunlem 2135  hbiu1 2588  onfr 2992  oarec 4202  zfregcl 4604  scott0 4727  cncnplem2 7772  chcmh 9108  atom1d 10275  fgsb 10555  fgsb2 10560

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 965  ax-4 975  ax-5o 977  ax-6o 980

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 983  df-rex 1653

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