Theorem poeq2 2857

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 2855	. . . 4 |- (A (_ B -> (R Po B -> R Po A))
2		poss 2855	. . . 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 2086	. . 3 |- (A = B <-> (A (_ B /\ B (_ A))
5		dfbi2 517	. . 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 524	1 |- (A = B -> (R Po A <-> R Po B))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223   = wceq 960   (_ wss 2056   Po wpo 2852

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 966  ax-gen 967  ax-8 968  ax-10 970  ax-12 972  ax-17 975  ax-4 977  ax-5o 979  ax-6o 982  ax-9o 1129  ax-10o 1146  ax-16 1216  ax-11o 1224  ax-ext 1466

This theorem depends on definitions:  df-bi 147  df-an 225  df-3an 781  df-ex 985  df-sb 1178  df-clab 1471  df-cleq 1476  df-clel 1479  df-ral 1656  df-in 2060  df-ss 2062  df-po 2854

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