Theorem nom60 337

Description: Part of Lemma 3.3(15) from "Non-Orthomodular Models..." paper. (Contributed by NM, 7-Feb-1999.)

Assertion

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

Proof of Theorem nom60

Step	Hyp	Ref	Expression
1		ancom 74	. . 3 ((b⊥ ∪ (a ∪ b)) ∩ ((a ∪ b)⊥ ∪ b)) = (((a ∪ b)⊥ ∪ b) ∩ (b⊥ ∪ (a ∪ b)))
2		df-id0 49	. . 3 (b ≡0 (a ∪ b)) = ((b⊥ ∪ (a ∪ b)) ∩ ((a ∪ b)⊥ ∪ b))
3		df-id0 49	. . 3 ((a ∪ b) ≡0 b) = (((a ∪ b)⊥ ∪ b) ∩ (b⊥ ∪ (a ∪ b)))
4	1, 2, 3	3tr1 63	. 2 (b ≡0 (a ∪ b)) = ((a ∪ b) ≡0 b)
5		nom50 331	. 2 ((a ∪ b) ≡0 b) = (a →2 b)
6	4, 5	ax-r2 36	1 (b ≡0 (a ∪ b)) = (a →2 b)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₂ wi2 13   ≡₀ wid0 17

This proof depends on 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 proof depends on definitions:  df-a 40  df-t 41  df-f 42  df-i1 44  df-i2 45  df-id0 49  df-le1 130  df-le2 131

This theorem is used by: (None)