Theorem nom30 319

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

Assertion

Ref	Expression
nom30	((a ∩ b) ≡0 a) = (a →1 b)

Proof of Theorem nom30

Step	Hyp	Ref	Expression
1		ancom 74	. . 3 (((a ∩ b)⊥ ∪ a) ∩ (a⊥ ∪ (a ∩ b))) = ((a⊥ ∪ (a ∩ b)) ∩ ((a ∩ b)⊥ ∪ a))
2		df-id0 49	. . 3 ((a ∩ b) ≡0 a) = (((a ∩ b)⊥ ∪ a) ∩ (a⊥ ∪ (a ∩ b)))
3		df-id0 49	. . 3 (a ≡0 (a ∩ b)) = ((a⊥ ∪ (a ∩ b)) ∩ ((a ∩ b)⊥ ∪ a))
4	1, 2, 3	3tr1 63	. 2 ((a ∩ b) ≡0 a) = (a ≡0 (a ∩ b))
5		nom20 313	. 2 (a ≡0 (a ∩ b)) = (a →1 b)
6	4, 5	ax-r2 36	1 ((a ∩ b) ≡0 a) = (a →1 b)

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12   ≡₀ wid0 17

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This theorem depends on definitions:  df-a 40  df-t 41  df-f 42  df-i1 44  df-id0 49  df-le1 130  df-le2 131

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