Theorem 2bi 99

Description: Join both sides with biconditional. (Contributed by NM, 10-Aug-1997.)

Hypotheses

Ref	Expression
2bi.1	a = b
2bi.2	c = d

Assertion

Ref	Expression
2bi	(a ≡ c) = (b ≡ d)

Proof of Theorem 2bi

Step	Hyp	Ref	Expression
1		2bi.2	. . 3 c = d
2	1	lbi 97	. 2 (a ≡ c) = (a ≡ d)
3		2bi.1	. . 3 a = b
4	3	rbi 98	. 2 (a ≡ d) = (b ≡ d)
5	2, 4	ax-r2 36	1 (a ≡ c) = (b ≡ d)

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