Theorem poeq2 2834

Description: Equality theorem for partial ordering predicate.

Assertion

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

Proof of Theorem poeq2

Step	Hyp	Ref	Expression
1		poss 2832	. . . 4 |- (A (_ B -> (R Po B -> R Po A))
2		poss 2832	. . . 4 |- (B (_ A -> (R Po A -> R Po B))
3	1, 2	anim12i 333	. . 3 |- ((A (_ B /\ B (_ A) -> ((R Po B -> R Po A) /\ (R Po A -> R Po B)))
4		eqss 2067	. . 3 |- (A = B <-> (A (_ B /\ B (_ A))
5		bi 513	. . 3 |- ((R Po B <-> R Po A) <-> ((R Po B -> R Po A) /\ (R Po A -> R Po B)))
6	3, 4, 5	3imtr4 219	. 2 |- (A = B -> (R Po B <-> R Po A))
7	6	bicomd 519	1 |- (A = B -> (R Po A <-> R Po B))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223   = wceq 953   (_ wss 2037   Po wpo 2829

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 959  ax-gen 960  ax-8 961  ax-10 963  ax-12 965  ax-17 968  ax-4 970  ax-5o 972  ax-6o 975  ax-9o 1119  ax-10o 1136  ax-16 1206  ax-11o 1213  ax-ext 1452

This theorem depends on definitions:  df-bi 147  df-an 225  df-3an 775  df-ex 978  df-sb 1168  df-clab 1457  df-cleq 1462  df-clel 1465  df-ral 1641  df-in 2041  df-ss 2043  df-po 2831

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