Theorem nom25 310

Description: Part of Lemma 3.3(14) from "Non-Orthomodular Models..." paper.

Assertion

Ref	Expression
nom25	(a ≡ (a ∩ b)) = (a →1 b)

Proof of Theorem nom25

Step	Hyp	Ref	Expression
1		anass 69	. . . . . 6 ((a ∩ a) ∩ b) = (a ∩ (a ∩ b))
2	1	ax-r1 34	. . . . 5 (a ∩ (a ∩ b)) = ((a ∩ a) ∩ b)
3		anidm 103	. . . . . 6 (a ∩ a) = a
4	3	ran 71	. . . . 5 ((a ∩ a) ∩ b) = (a ∩ b)
5	2, 4	ax-r2 35	. . . 4 (a ∩ (a ∩ b)) = (a ∩ b)
6		oran3 85	. . . . . . 7 (a⊥ ∪ b⊥ ) = (a ∩ b)⊥
7	6	lan 70	. . . . . 6 (a⊥ ∩ (a⊥ ∪ b⊥ )) = (a⊥ ∩ (a ∩ b)⊥ )
8	7	ax-r1 34	. . . . 5 (a⊥ ∩ (a ∩ b)⊥ ) = (a⊥ ∩ (a⊥ ∪ b⊥ ))
9		a5c 113	. . . . 5 (a⊥ ∩ (a⊥ ∪ b⊥ )) = a⊥
10	8, 9	ax-r2 35	. . . 4 (a⊥ ∩ (a ∩ b)⊥ ) = a⊥
11	5, 10	2or 67	. . 3 ((a ∩ (a ∩ b)) ∪ (a⊥ ∩ (a ∩ b)⊥ )) = ((a ∩ b) ∪ a⊥ )
12		ax-a2 30	. . 3 ((a ∩ b) ∪ a⊥ ) = (a⊥ ∪ (a ∩ b))
13	11, 12	ax-r2 35	. 2 ((a ∩ (a ∩ b)) ∪ (a⊥ ∩ (a ∩ b)⊥ )) = (a⊥ ∪ (a ∩ b))
14		dfb 86	. 2 (a ≡ (a ∩ b)) = ((a ∩ (a ∩ b)) ∪ (a⊥ ∩ (a ∩ b)⊥ ))
15		df-i1 43	. 2 (a →1 b) = (a⊥ ∪ (a ∩ b))
16	13, 14, 15	3tr1 60	1 (a ≡ (a ∩ b)) = (a →1 b)

Syntax hints:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7   →₁ wi1 13

This theorem is referenced by:  nom35 316  nom55 328

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37

This theorem depends on definitions:  df-b 38  df-a 39  df-t 40  df-f 41  df-i1 43

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