Theorem freq1 2174

Description: Equality theorem for the founded predicate.

Assertion

Ref	Expression
freq1	|- (R = S -> (R Fr A <-> S Fr A))

Proof of Theorem freq1

Step	Hyp	Ref	Expression
1		breq 2064	. . . . . . 7 |- (R = S -> (zRy <-> zSy))
2	1	negbid 463	. . . . . 6 |- (R = S -> (-. zRy <-> -. zSy))
3	2	biraldv 1219	. . . . 5 |- (R = S -> (A.z e. x -. zRy <-> A.z e. x -. zSy))
4	3	birexdv 1220	. . . 4 |- (R = S -> (E.y e. x A.z e. x -. zRy <-> E.y e. x A.z e. x -. zSy))
5	4	imbi2d 464	. . 3 |- (R = S -> (((x (_ A /\ -. x = (/)) -> E.y e. x A.z e. x -. zRy) <-> ((x (_ A /\ -. x = (/)) -> E.y e. x A.z e. x -. zSy)))
6	5	bialdv 935	. 2 |- (R = S -> (A.x((x (_ A /\ -. x = (/)) -> E.y e. x A.z e. x -. zRy) <-> A.x((x (_ A /\ -. x = (/)) -> E.y e. x A.z e. x -. zSy)))
7		df-fr 2169	. 2 |- (R Fr A <-> A.x((x (_ A /\ -. x = (/)) -> E.y e. x A.z e. x -. zRy))
8		df-fr 2169	. 2 |- (S Fr A <-> A.x((x (_ A /\ -. x = (/)) -> E.y e. x A.z e. x -. zSy))
9	6, 7, 8	3bitr4g 428	1 |- (R = S -> (R Fr A <-> S Fr A))

Syntax hints:  -. wn 1   -> wi 2   <-> wb 127   /\ wa 196  A.wal 672   = wceq 1091  A.wral 1201  E.wrex 1202   (_ wss 1487  (/)c0 1707   class class class wbr 2054   Fr wfr 2061

This theorem is referenced by:  weeq1 2189

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-gen 677  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-cleq 1097  df-clel 1099  df-ral 1205  df-rex 1206  df-br 2063  df-fr 2169

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