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Theorem 2bi 99
Description: Join both sides with biconditional.
Hypotheses
Ref Expression
2bi.1 a = b
2bi.2 c = d
Assertion
Ref Expression
2bi (a == c) = (b == d)
Proof of Theorem 2bi
StepHypRef Expression
1 2bi.2 . . 3 c = d
21lbi 97 . 2 (a == c) = (a == d)
3 2bi.1 . . 3 a = b
43rbi 98 . 2 (a == d) = (b == d)
52, 4ax-r2 36 1 (a == c) = (b == d)
Colors of variables: term
Syntax hints:   = wb 1   == tb 5
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:  wwfh3  218  wwfh4  219  ska2a  226  ska2b  227  ka4lem  229  wlor  368  wran  369  wlan  370  wom2  434  u3lemax4  796  mlaconj4  844
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