Theorem 2bi 91

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 89	. 2 (a ≡ c) = (a ≡ d)
3		2bi.1	. . 3 a = b
4	3	rbi 90	. 2 (a ≡ d) = (b ≡ d)
5	2, 4	ax-r2 35	1 (a ≡ c) = (b ≡ d)

Syntax hints:   = wb 1   ≡ tb 5

This theorem is referenced by:  wwfh3 210  wwfh4 211  ska2a 218  ska2b 219  ka4lem 221  wlor 350  wran 351  wlan 352  wom2 416  u3lemax4 778  mlaconj4 826

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37

This theorem depends on definitions:  df-b 38  df-a 39

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