Theorem nom50 323

Description: Part of Lemma 3.3(15) from "Non-Orthomodular Models..." paper.

Assertion

Ref	Expression
nom50	((a ∪ b) ≡0 b) = (a →2 b)

Proof of Theorem nom50

Step	Hyp	Ref	Expression
1		ancom 68	. . . . . . . 8 (b⊥ ∩ a⊥ ) = (a⊥ ∩ b⊥ )
2		anor3 82	. . . . . . . 8 (a⊥ ∩ b⊥ ) = (a ∪ b)⊥
3	1, 2	ax-r2 35	. . . . . . 7 (b⊥ ∩ a⊥ ) = (a ∪ b)⊥
4	3	lor 66	. . . . . 6 (b⊥ ⊥ ∪ (b⊥ ∩ a⊥ )) = (b⊥ ⊥ ∪ (a ∪ b)⊥ )
5	3	ax-r4 36	. . . . . . 7 (b⊥ ∩ a⊥ )⊥ = (a ∪ b)⊥ ⊥
6	5	ax-r5 37	. . . . . 6 ((b⊥ ∩ a⊥ )⊥ ∪ b⊥ ) = ((a ∪ b)⊥ ⊥ ∪ b⊥ )
7	4, 6	2an 72	. . . . 5 ((b⊥ ⊥ ∪ (b⊥ ∩ a⊥ )) ∩ ((b⊥ ∩ a⊥ )⊥ ∪ b⊥ )) = ((b⊥ ⊥ ∪ (a ∪ b)⊥ ) ∩ ((a ∪ b)⊥ ⊥ ∪ b⊥ ))
8	7	ax-r1 34	. . . 4 ((b⊥ ⊥ ∪ (a ∪ b)⊥ ) ∩ ((a ∪ b)⊥ ⊥ ∪ b⊥ )) = ((b⊥ ⊥ ∪ (b⊥ ∩ a⊥ )) ∩ ((b⊥ ∩ a⊥ )⊥ ∪ b⊥ ))
9		df-id0 48	. . . 4 (b⊥ ≡0 (a ∪ b)⊥ ) = ((b⊥ ⊥ ∪ (a ∪ b)⊥ ) ∩ ((a ∪ b)⊥ ⊥ ∪ b⊥ ))
10		df-id0 48	. . . 4 (b⊥ ≡0 (b⊥ ∩ a⊥ )) = ((b⊥ ⊥ ∪ (b⊥ ∩ a⊥ )) ∩ ((b⊥ ∩ a⊥ )⊥ ∪ b⊥ ))
11	8, 9, 10	3tr1 60	. . 3 (b⊥ ≡0 (a ∪ b)⊥ ) = (b⊥ ≡0 (b⊥ ∩ a⊥ ))
12		nom20 305	. . 3 (b⊥ ≡0 (b⊥ ∩ a⊥ )) = (b⊥ →1 a⊥ )
13	11, 12	ax-r2 35	. 2 (b⊥ ≡0 (a ∪ b)⊥ ) = (b⊥ →1 a⊥ )
14		nomcon0 293	. 2 ((a ∪ b) ≡0 b) = (b⊥ ≡0 (a ∪ b)⊥ )
15		i2i1 259	. 2 (a →2 b) = (b⊥ →1 a⊥ )
16	13, 14, 15	3tr1 60	1 ((a ∪ b) ≡0 b) = (a →2 b)

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 13   →₂ wi2 14   ≡₀ wid0 18

This theorem is referenced by:  nom60 329

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-a 39  df-t 40  df-f 41  df-i1 43  df-i2 44  df-id0 48  df-le1 122  df-le2 123

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