Theorem lbi 89

Description: Introduce biconditional to the left.

Hypothesis

Ref	Expression
lbi.1	a = b

Assertion

Ref	Expression
lbi	(c ≡ a) = (c ≡ b)

Proof of Theorem lbi

Step	Hyp	Ref	Expression
1		lbi.1	. . . 4 a = b
2	1	lan 70	. . 3 (c ∩ a) = (c ∩ b)
3	1	ax-r4 36	. . . 4 a⊥ = b⊥
4	3	lan 70	. . 3 (c⊥ ∩ a⊥ ) = (c⊥ ∩ b⊥ )
5	2, 4	2or 67	. 2 ((c ∩ a) ∪ (c⊥ ∩ a⊥ )) = ((c ∩ b) ∪ (c⊥ ∩ b⊥ ))
6		dfb 86	. 2 (c ≡ a) = ((c ∩ a) ∪ (c⊥ ∩ a⊥ ))
7		dfb 86	. 2 (c ≡ b) = ((c ∩ b) ∪ (c⊥ ∩ b⊥ ))
8	5, 6, 7	3tr1 60	1 (c ≡ a) = (c ≡ b)

Syntax hints:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7

This theorem is referenced by:  rbi 90  2bi 91  wcon3 201  wwoml2 204  nom55 328

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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