Theorem poeq2 2849

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 2847	. . . 4 |- (A (_ B -> (R Po B -> R Po A))
2		poss 2847	. . . 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 2080	. . 3 |- (A = B <-> (A (_ B /\ B (_ A))
5		dfbi2 516	. . 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 523	1 |- (A = B -> (R Po A <-> R Po B))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223   = wceq 958   (_ wss 2050   Po wpo 2844

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-12 970  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-16 1212  ax-11o 1220  ax-ext 1462

This theorem depends on definitions:  df-bi 147  df-an 225  df-3an 779  df-ex 983  df-sb 1174  df-clab 1467  df-cleq 1472  df-clel 1475  df-ral 1652  df-in 2054  df-ss 2056  df-po 2846

Copyright terms: Public domain