Theorem nom60 329

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

Assertion

Ref	Expression
nom60	(b ≡0 (a ∪ b)) = (a →2 b)

Proof of Theorem nom60

Step	Hyp	Ref	Expression
1		ancom 68	. . 3 ((b⊥ ∪ (a ∪ b)) ∩ ((a ∪ b)⊥ ∪ b)) = (((a ∪ b)⊥ ∪ b) ∩ (b⊥ ∪ (a ∪ b)))
2		df-id0 48	. . 3 (b ≡0 (a ∪ b)) = ((b⊥ ∪ (a ∪ b)) ∩ ((a ∪ b)⊥ ∪ b))
3		df-id0 48	. . 3 ((a ∪ b) ≡0 b) = (((a ∪ b)⊥ ∪ b) ∩ (b⊥ ∪ (a ∪ b)))
4	1, 2, 3	3tr1 60	. 2 (b ≡0 (a ∪ b)) = ((a ∪ b) ≡0 b)
5		nom50 323	. 2 ((a ∪ b) ≡0 b) = (a →2 b)
6	4, 5	ax-r2 35	1 (b ≡0 (a ∪ b)) = (a →2 b)

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

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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