Theorem hbreu1 1765

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

Assertion

Ref	Expression
hbreu1	|- (E!x e. A ph -> A.xE!x e. A ph)

Proof of Theorem hbreu1

Step	Hyp	Ref	Expression
1		hbeu1 1386	. 2 |- (E!x(x e. A /\ ph) -> A.xE!x(x e. A /\ ph))
2		df-reu 1648	. 2 |- (E!x e. A ph <-> E!x(x e. A /\ ph))
3	2	albii 997	. 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 952   e. wcel 956  E!weu 1378  E!wreu 1644

This theorem is referenced by:  reuuni2f 2878  reuuni4 2882  isumclt 7152

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-4 971  ax-5o 973  ax-6o 976

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 979  df-eu 1380  df-reu 1648

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