Theorem poeq1 2842

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 2621	. . . . . 6 |- (R = S -> (xRx <-> xSx))
2	1	negbid 611	. . . . 5 |- (R = S -> (-. xRx <-> -. xSx))
3		breq 2621	. . . . . . 7 |- (R = S -> (xRy <-> xSy))
4		breq 2621	. . . . . . 7 |- (R = S -> (yRz <-> ySz))
5	3, 4	anbi12d 628	. . . . . 6 |- (R = S -> ((xRy /\ yRz) <-> (xSy /\ ySz)))
6		breq 2621	. . . . . 6 |- (R = S -> (xRz <-> xSz))
7	5, 6	imbi12d 626	. . . . 5 |- (R = S -> (((xRy /\ yRz) -> xRz) <-> ((xSy /\ ySz) -> xSz)))
8	2, 7	anbi12d 628	. . . 4 |- (R = S -> ((-. xRx /\ ((xRy /\ yRz) -> xRz)) <-> (-. xSx /\ ((xSy /\ ySz) -> xSz))))
9	8	ralbidv 1663	. . 3 |- (R = S -> (A.z e. A (-. xRx /\ ((xRy /\ yRz) -> xRz)) <-> A.z e. A (-. xSx /\ ((xSy /\ ySz) -> xSz))))
10	9	2ralbidv 1680	. 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 2840	. 2 |- (R Po A <-> A.x e. A A.y e. A A.z e. A (-. xRx /\ ((xRy /\ yRz) -> xRz)))
12		df-po 2840	. 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 555	1 |- (R = S -> (R Po A <-> S Po A))

Syntax hints:  -. wn 2   -> wi 3   <-> wb 146   /\ wa 223   = wceq 956  A.wral 1645   class class class wbr 2619   Po wpo 2838

This theorem is referenced by:  soeq1 2853

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 963  ax-17 971  ax-4 973  ax-5o 975  ax-ext 1459

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-cleq 1469  df-clel 1472  df-ral 1649  df-br 2620  df-po 2840

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