Theorem poeq2 4365

Description: Equality theorem for partial ordering predicate. (Contributed by NM, 27-Mar-1997.)

Assertion

Ref	Expression
poeq2	|- ( A = B -> ( R Po A <-> R Po B ) )

Proof of Theorem poeq2

Step	Hyp	Ref	Expression
1		eqimss2 3256	. . 3 |- ( A = B -> B C_ A )
2		poss 4363	. . 3 |- ( B C_ A -> ( R Po A -> R Po B ) )
3	1, 2	syl 14	. 2 |- ( A = B -> ( R Po A -> R Po B ) )
4		eqimss 3255	. . 3 |- ( A = B -> A C_ B )
5		poss 4363	. . 3 |- ( A C_ B -> ( R Po B -> R Po A ) )
6	4, 5	syl 14	. 2 |- ( A = B -> ( R Po B -> R Po A ) )
7	3, 6	impbid 129	1 |- ( A = B -> ( R Po A <-> R Po B ) )

Syntax hints:   -> wi 4   <-> wb 105   = wceq 1373   C_ wss 3174   Po wpo 4359

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-11 1530  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2189

This theorem depends on definitions:  df-bi 117  df-nf 1485  df-sb 1787  df-clab 2194  df-cleq 2200  df-clel 2203  df-ral 2491  df-in 3180  df-ss 3187  df-po 4361

This theorem is referenced by: (None)