Theorem hbre1 1687

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 1015	. 2 |- (E.x(x e. A /\ ph) -> A.xE.x(x e. A /\ ph))
2		df-rex 1648	. 2 |- (E.x e. A ph <-> E.x(x e. A /\ ph))
3	2	albii 998	. 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 953   e. wcel 957  E.wex 979  E.wrex 1644

This theorem is referenced by:  uniiunlem 2129  hbiu1 2580  onfr 2982  oarec 4189  iunfi 4552  zfregcl 4578  scott0 4700  cncnplem2 7735  chcmh 9068  atom1d 10236  fgsb 10503  fgsb2 10508

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 962  ax-4 972  ax-5o 974  ax-6o 977

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 980  df-rex 1648

Copyright terms: Public domain