Theorem elabg 1890

Description: Membership in a class abstraction with implicit substitution. Compare Theorem 6.13 of [Quine] p. 44.

Hypothesis

Ref	Expression
elabg.1	|- (x = A -> (ph <-> ps))

Assertion

Ref	Expression
elabg	|- (A e. B -> (A e. {x | ph} <-> ps))

Distinct variable groups:   ps,x   x,A

Proof of Theorem elabg

Step	Hyp	Ref	Expression
1		ax-17 968	. 2 |- (y e. A -> A.x y e. A)
2		ax-17 968	. 2 |- (ps -> A.xps)
3		elabg.1	. 2 |- (x = A -> (ph <-> ps))
4	1, 2, 3	elabgf 1889	1 |- (A e. B -> (A e. {x | ph} <-> ps))

Syntax hints:   -> wi 3   <-> wb 146   = wceq 953   e. wcel 955  {cab 1456

This theorem is referenced by:  elab2g 1891  elab3g 1893  intmin3 2548  finds 3146  scott0 4689  elcncf 7200  eltgt 7560  eltg2t 7561  iscld 7611  dfpjopt 10021  spfi 10346  ishomeo 10404  eloi 10503  ismonb 10580  isfunb 10593

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 959  ax-gen 960  ax-8 961  ax-10 963  ax-12 965  ax-17 968  ax-4 970  ax-5o 972  ax-6o 975  ax-9o 1119  ax-10o 1136  ax-16 1206  ax-11o 1213  ax-ext 1452

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 978  df-sb 1168  df-clab 1457  df-cleq 1462  df-clel 1465  df-v 1803

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