Theorem poeq1 2837

Description: Equality theorem for partial ordering predicate.

Assertion

Ref	Expression
poeq1	|- (R = S -> (R Po A <-> S Po A))

Proof of Theorem poeq1

Step	Hyp	Ref	Expression
1		breq 2616	. . . . . 6 |- (R = S -> (xRx <-> xSx))
2	1	negbid 610	. . . . 5 |- (R = S -> (-. xRx <-> -. xSx))
3		breq 2616	. . . . . . 7 |- (R = S -> (xRy <-> xSy))
4		breq 2616	. . . . . . 7 |- (R = S -> (yRz <-> ySz))
5	3, 4	anbi12d 627	. . . . . 6 |- (R = S -> ((xRy /\ yRz) <-> (xSy /\ ySz)))
6		breq 2616	. . . . . 6 |- (R = S -> (xRz <-> xSz))
7	5, 6	imbi12d 625	. . . . 5 |- (R = S -> (((xRy /\ yRz) -> xRz) <-> ((xSy /\ ySz) -> xSz)))
8	2, 7	anbi12d 627	. . . 4 |- (R = S -> ((-. xRx /\ ((xRy /\ yRz) -> xRz)) <-> (-. xSx /\ ((xSy /\ ySz) -> xSz))))
9	8	ralbidv 1660	. . 3 |- (R = S -> (A.z e. A (-. xRx /\ ((xRy /\ yRz) -> xRz)) <-> A.z e. A (-. xSx /\ ((xSy /\ ySz) -> xSz))))
10	9	2ralbidv 1677	. 2 |- (R = S -> (A.x e. A A.y e. A A.z e. A (-. xRx /\ ((xRy /\ yRz) -> xRz)) <-> A.x e. A A.y e. A A.z e. A (-. xSx /\ ((xSy /\ ySz) -> xSz))))
11		df-po 2835	. 2 |- (R Po A <-> A.x e. A A.y e. A A.z e. A (-. xRx /\ ((xRy /\ yRz) -> xRz)))
12		df-po 2835	. 2 |- (S Po A <-> A.x e. A A.y e. A A.z e. A (-. xSx /\ ((xSy /\ ySz) -> xSz)))
13	10, 11, 12	3bitr4g 554	1 |- (R = S -> (R Po A <-> S Po A))

Syntax hints:  -. wn 2   -> wi 3   <-> wb 146   /\ wa 223   = wceq 954  A.wral 1642   class class class wbr 2614   Po wpo 2833

This theorem is referenced by:  soeq1 2848

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-br 2615  df-po 2835

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