Theorem poeq2 2131

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 2129	. . . 4 |- (A (_ B -> (R Po B -> R Po A))
2		poss 2129	. . . 4 |- (B (_ A -> (R Po A -> R Po B))
3	1, 2	anim12i 268	. . 3 |- ((A (_ B /\ B (_ A) -> ((R Po B -> R Po A) /\ (R Po A -> R Po B)))
4		eqss 1516	. . 3 |- (A = B <-> (A (_ B /\ B (_ A))
5		bi 396	. . 3 |- ((R Po B <-> R Po A) <-> ((R Po B -> R Po A) /\ (R Po A -> R Po B)))
6	3, 4, 5	3imtr4 192	. 2 |- (A = B -> (R Po B <-> R Po A))
7	6	bicomd 399	1 |- (A = B -> (R Po A <-> R Po B))

Syntax hints:   -> wi 2   <-> wb 127   /\ wa 196   = wceq 1091   (_ wss 1487   Po wpo 2058

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-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-3an 583  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-ral 1205  df-in 1491  df-ss 1492  df-po 2128

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