Theorem biid 116

Description: Identity law.

Assertion

Ref	Expression
biid	(a ≡ a) = 1

Proof of Theorem biid

Step	Hyp	Ref	Expression
1		anidm 111	. . 3 (a ∩ a) = a
2		anidm 111	. . 3 (a⊥ ∩ a⊥ ) = a⊥
3	1, 2	2or 72	. 2 ((a ∩ a) ∪ (a⊥ ∩ a⊥ )) = (a ∪ a⊥ )
4		dfb 94	. 2 (a ≡ a) = ((a ∩ a) ∪ (a⊥ ∩ a⊥ ))
5		df-t 41	. 2 1 = (a ∪ a⊥ )
6	3, 4, 5	3tr1 63	1 (a ≡ a) = 1

Syntax hints:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7  1wt 8

This theorem is referenced by:  bi1 118  ska1 231  wdid0id1 1109

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42

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