Theorem soeq1 2141

Description: Equality theorem for the strict ordering predicate.

Assertion

Ref	Expression
soeq1	|- (R = S -> (R Or A <-> S Or A))

Proof of Theorem soeq1

Step	Hyp	Ref	Expression
1		poeq1 2130	. . 3 |- (R = S -> (R Po A <-> S Po A))
2		breq 2064	. . . . . 6 |- (R = S -> (xRy <-> xSy))
3		pm4.2i 149	. . . . . 6 |- (R = S -> (x = y <-> x = y))
4		breq 2064	. . . . . 6 |- (R = S -> (yRx <-> ySx))
5	2, 3, 4	bi3ord 635	. . . . 5 |- (R = S -> ((xRy \/ x = y \/ yRx) <-> (xSy \/ x = y \/ ySx)))
6	5	biraldv 1219	. . . 4 |- (R = S -> (A.y e. A (xRy \/ x = y \/ yRx) <-> A.y e. A (xSy \/ x = y \/ ySx)))
7	6	biraldv 1219	. . 3 |- (R = S -> (A.x e. A A.y e. A (xRy \/ x = y \/ yRx) <-> A.x e. A A.y e. A (xSy \/ x = y \/ ySx)))
8	1, 7	anbi12d 476	. 2 |- (R = S -> ((R Po A /\ A.x e. A A.y e. A (xRy \/ x = y \/ yRx)) <-> (S Po A /\ A.x e. A A.y e. A (xSy \/ x = y \/ ySx))))
9		df-so 2138	. 2 |- (R Or A <-> (R Po A /\ A.x e. A A.y e. A (xRy \/ x = y \/ yRx)))
10		df-so 2138	. 2 |- (S Or A <-> (S Po A /\ A.x e. A A.y e. A (xSy \/ x = y \/ ySx)))
11	8, 9, 10	3bitr4g 428	1 |- (R = S -> (R Or A <-> S Or A))

Syntax hints:   -> wi 2   <-> wb 127   /\ wa 196   \/ w3o 580   = weq 797   = wceq 1091  A.wral 1201   class class class wbr 2054   Po wpo 2058   Or wor 2059

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-or 197  df-an 198  df-3or 582  df-ex 679  df-cleq 1097  df-clel 1099  df-ral 1205  df-br 2063  df-po 2128  df-so 2138

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