Theorem poeq1 2130

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 2064	. . . . . . 7 |- (R = S -> (xRx <-> xSx))
2	1	negbid 463	. . . . . 6 |- (R = S -> (-. xRx <-> -. xSx))
3		breq 2064	. . . . . . . 8 |- (R = S -> (xRy <-> xSy))
4		breq 2064	. . . . . . . 8 |- (R = S -> (yRz <-> ySz))
5	3, 4	anbi12d 476	. . . . . . 7 |- (R = S -> ((xRy /\ yRz) <-> (xSy /\ ySz)))
6		breq 2064	. . . . . . 7 |- (R = S -> (xRz <-> xSz))
7	5, 6	imbi12d 474	. . . . . 6 |- (R = S -> (((xRy /\ yRz) -> xRz) <-> ((xSy /\ ySz) -> xSz)))
8	2, 7	anbi12d 476	. . . . 5 |- (R = S -> ((-. xRx /\ ((xRy /\ yRz) -> xRz)) <-> (-. xSx /\ ((xSy /\ ySz) -> xSz))))
9	8	biraldv 1219	. . . 4 |- (R = S -> (A.z e. A (-. xRx /\ ((xRy /\ yRz) -> xRz)) <-> A.z e. A (-. xSx /\ ((xSy /\ ySz) -> xSz))))
10	9	biraldv 1219	. . 3 |- (R = S -> (A.y e. A A.z e. A (-. xRx /\ ((xRy /\ yRz) -> xRz)) <-> A.y e. A A.z e. A (-. xSx /\ ((xSy /\ ySz) -> xSz))))
11	10	biraldv 1219	. 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))))
12		df-po 2128	. 2 |- (R Po A <-> A.x e. A A.y e. A A.z e. A (-. xRx /\ ((xRy /\ yRz) -> xRz)))
13		df-po 2128	. 2 |- (S Po A <-> A.x e. A A.y e. A A.z e. A (-. xSx /\ ((xSy /\ ySz) -> xSz)))
14	11, 12, 13	3bitr4g 428	1 |- (R = S -> (R Po A <-> S Po A))

Syntax hints:  -. wn 1   -> wi 2   <-> wb 127   /\ wa 196   = wceq 1091  A.wral 1201   class class class wbr 2054   Po wpo 2058

This theorem is referenced by:  soeq1 2141

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-br 2063  df-po 2128

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