Theorem nom40 317

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

Assertion

Ref	Expression
nom40	((a ∪ b) →0 b) = (a →2 b)

Proof of Theorem nom40

Step	Hyp	Ref	Expression
1		nom10 299	. 2 (b⊥ →0 (b⊥ ∩ a⊥ )) = (b⊥ →1 a⊥ )
2		ax-a2 30	. . . 4 ((a ∪ b)⊥ ∪ b) = (b ∪ (a ∪ b)⊥ )
3		ax-a1 29	. . . . 5 b = b⊥ ⊥
4		ancom 68	. . . . . . 7 (b⊥ ∩ a⊥ ) = (a⊥ ∩ b⊥ )
5		anor3 82	. . . . . . 7 (a⊥ ∩ b⊥ ) = (a ∪ b)⊥
6	4, 5	ax-r2 35	. . . . . 6 (b⊥ ∩ a⊥ ) = (a ∪ b)⊥
7	6	ax-r1 34	. . . . 5 (a ∪ b)⊥ = (b⊥ ∩ a⊥ )
8	3, 7	2or 67	. . . 4 (b ∪ (a ∪ b)⊥ ) = (b⊥ ⊥ ∪ (b⊥ ∩ a⊥ ))
9	2, 8	ax-r2 35	. . 3 ((a ∪ b)⊥ ∪ b) = (b⊥ ⊥ ∪ (b⊥ ∩ a⊥ ))
10		df-i0 42	. . 3 ((a ∪ b) →0 b) = ((a ∪ b)⊥ ∪ b)
11		df-i0 42	. . 3 (b⊥ →0 (b⊥ ∩ a⊥ )) = (b⊥ ⊥ ∪ (b⊥ ∩ a⊥ ))
12	9, 10, 11	3tr1 60	. 2 ((a ∪ b) →0 b) = (b⊥ →0 (b⊥ ∩ a⊥ ))
13		i2i1 259	. 2 (a →2 b) = (b⊥ →1 a⊥ )
14	1, 12, 13	3tr1 60	1 ((a ∪ b) →0 b) = (a →2 b)

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₀ wi0 12   →₁ wi1 13   →₂ wi2 14

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-a 39  df-i0 42  df-i1 43  df-i2 44

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