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

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 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

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

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