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Theorem lbi 97
Description: Introduce biconditional to the left. (Contributed by NM, 10-Aug-1997.)
Hypothesis
Ref Expression
lbi.1 a = b
Assertion
Ref Expression
lbi (ca) = (cb)

Proof of Theorem lbi
StepHypRef Expression
1 lbi.1 . . . 4 a = b
21lan 77 . . 3 (ca) = (cb)
31ax-r4 37 . . . 4 a = b
43lan 77 . . 3 (ca ) = (cb )
52, 42or 72 . 2 ((ca) ∪ (ca )) = ((cb) ∪ (cb ))
6 dfb 94 . 2 (ca) = ((ca) ∪ (ca ))
7 dfb 94 . 2 (cb) = ((cb) ∪ (cb ))
85, 6, 73tr1 63 1 (ca) = (cb)
Colors of variables: term
Syntax hints:   = wb 1   wn 4  tb 5  wo 6  wa 7
This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38
This theorem depends on definitions:  df-b 39  df-a 40
This theorem is referenced by:  rbi  98  2bi  99  wcon3  209  wwoml2  212  nom55  336
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