Theorem freq1 2917

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 2616	. . . . . 6 |- (R = S -> (zRy <-> zSy))
2	1	negbid 610	. . . . 5 |- (R = S -> (-. zRy <-> -. zSy))
3	2	rexralbidv 1679	. . . 4 |- (R = S -> (E.y e. x A.z e. x -. zRy <-> E.y e. x A.z e. x -. zSy))
4	3	imbi2d 611	. . 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)))
5	4	albidv 1276	. 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)))
6		df-fr 2912	. 2 |- (R Fr A <-> A.x((x (_ A /\ x =/= (/)) -> E.y e. x A.z e. x -. zRy))
7		df-fr 2912	. 2 |- (S Fr A <-> A.x((x (_ A /\ x =/= (/)) -> E.y e. x A.z e. x -. zSy))
8	5, 6, 7	3bitr4g 554	1 |- (R = S -> (R Fr A <-> S Fr A))

Syntax hints:  -. wn 2   -> wi 3   <-> wb 146   /\ wa 223  A.wal 952   = wceq 954   =/= wne 1582  A.wral 1642  E.wrex 1643   (_ wss 2043  (/)c0 2276   class class class wbr 2614   Fr wfr 2910

This theorem is referenced by:  weeq1 2932

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 961  ax-17 969  ax-4 971  ax-5o 973  ax-ext 1457

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 979  df-cleq 1467  df-clel 1470  df-ral 1646  df-rex 1647  df-br 2615  df-fr 2912

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