Theorem nom40 325

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

Assertion

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

Proof of Theorem nom40

Step	Hyp	Ref	Expression
1		nom10 307	. 2 (b⊥ →0 (b⊥ ∩ a⊥ )) = (b⊥ →1 a⊥ )
2		ax-a2 31	. . . 4 ((a ∪ b)⊥ ∪ b) = (b ∪ (a ∪ b)⊥ )
3		ax-a1 30	. . . . 5 b = b⊥ ⊥
4		ancom 74	. . . . . . 7 (b⊥ ∩ a⊥ ) = (a⊥ ∩ b⊥ )
5		anor3 90	. . . . . . 7 (a⊥ ∩ b⊥ ) = (a ∪ b)⊥
6	4, 5	ax-r2 36	. . . . . 6 (b⊥ ∩ a⊥ ) = (a ∪ b)⊥
7	6	ax-r1 35	. . . . 5 (a ∪ b)⊥ = (b⊥ ∩ a⊥ )
8	3, 7	2or 72	. . . 4 (b ∪ (a ∪ b)⊥ ) = (b⊥ ⊥ ∪ (b⊥ ∩ a⊥ ))
9	2, 8	ax-r2 36	. . 3 ((a ∪ b)⊥ ∪ b) = (b⊥ ⊥ ∪ (b⊥ ∩ a⊥ ))
10		df-i0 43	. . 3 ((a ∪ b) →0 b) = ((a ∪ b)⊥ ∪ b)
11		df-i0 43	. . 3 (b⊥ →0 (b⊥ ∩ a⊥ )) = (b⊥ ⊥ ∪ (b⊥ ∩ a⊥ ))
12	9, 10, 11	3tr1 63	. 2 ((a ∪ b) →0 b) = (b⊥ →0 (b⊥ ∩ a⊥ ))
13		i2i1 267	. 2 (a →2 b) = (b⊥ →1 a⊥ )
14	1, 12, 13	3tr1 63	1 ((a ∪ b) →0 b) = (a →2 b)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₀ wi0 11   →₁ wi1 12   →₂ wi2 13

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This proof depends on definitions:  df-a 40  df-i0 43  df-i1 44  df-i2 45

This theorem is used by: (None)