Theorem soeq1 5054

Description: Equality theorem for the strict ordering predicate. (Contributed by NM, 16-Mar-1997.)

Assertion

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

Proof of Theorem soeq1

Dummy variables x y are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		poeq1 5038	. . 3 |- ( R = S -> ( R Po A <-> S Po A ) )
2		breq 4655	. . . . 5 |- ( R = S -> ( x R y <-> x S y ) )
3		biidd 252	. . . . 5 |- ( R = S -> ( x = y <-> x = y ) )
4		breq 4655	. . . . 5 |- ( R = S -> ( y R x <-> y S x ) )
5	2, 3, 4	3orbi123d 1398	. . . 4 |- ( R = S -> ( ( x R y \/ x = y \/ y R x ) <-> ( x S y \/ x = y \/ y S x ) ) )
6	5	2ralbidv 2989	. . 3 |- ( R = S -> ( A. x e. A A. y e. A ( x R y \/ x = y \/ y R x ) <-> A. x e. A A. y e. A ( x S y \/ x = y \/ y S x ) ) )
7	1, 6	anbi12d 747	. 2 |- ( R = S -> ( ( R Po A /\ A. x e. A A. y e. A ( x R y \/ x = y \/ y R x ) ) <-> ( S Po A /\ A. x e. A A. y e. A ( x S y \/ x = y \/ y S x ) ) ) )
8		df-so 5036	. 2 |- ( R Or A <-> ( R Po A /\ A. x e. A A. y e. A ( x R y \/ x = y \/ y R x ) ) )
9		df-so 5036	. 2 |- ( S Or A <-> ( S Po A /\ A. x e. A A. y e. A ( x S y \/ x = y \/ y S x ) ) )
10	7, 8, 9	3bitr4g 303	1 |- ( R = S -> ( R Or A <-> S Or A ) )

Syntax hints:   -> wi 4   <-> wb 196   /\ wa 384   \/ w3o 1036   = wceq 1483   A.wral 2912   class class class wbr 4653   Po wpo 5033   Or wor 5034

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3or 1038  df-ex 1705  df-cleq 2615  df-clel 2618  df-ral 2917  df-br 4654  df-po 5035  df-so 5036

This theorem is referenced by:  weeq1  5102  ltsopi  9710  cnso  14976  opsrtoslem2  19485  soeq12d  37608