Theorem nom25 318

Description: Part of Lemma 3.3(14) from "Non-Orthomodular Models..." paper. (Contributed by NM, 7-Feb-1999.)

Assertion

Ref	Expression
nom25	(a ≡ (a ∩ b)) = (a →1 b)

Proof of Theorem nom25

Step	Hyp	Ref	Expression
1		anass 76	. . . . . 6 ((a ∩ a) ∩ b) = (a ∩ (a ∩ b))
2	1	ax-r1 35	. . . . 5 (a ∩ (a ∩ b)) = ((a ∩ a) ∩ b)
3		anidm 111	. . . . . 6 (a ∩ a) = a
4	3	ran 78	. . . . 5 ((a ∩ a) ∩ b) = (a ∩ b)
5	2, 4	ax-r2 36	. . . 4 (a ∩ (a ∩ b)) = (a ∩ b)
6		oran3 93	. . . . . . 7 (a⊥ ∪ b⊥ ) = (a ∩ b)⊥
7	6	lan 77	. . . . . 6 (a⊥ ∩ (a⊥ ∪ b⊥ )) = (a⊥ ∩ (a ∩ b)⊥ )
8	7	ax-r1 35	. . . . 5 (a⊥ ∩ (a ∩ b)⊥ ) = (a⊥ ∩ (a⊥ ∪ b⊥ ))
9		anabs 121	. . . . 5 (a⊥ ∩ (a⊥ ∪ b⊥ )) = a⊥
10	8, 9	ax-r2 36	. . . 4 (a⊥ ∩ (a ∩ b)⊥ ) = a⊥
11	5, 10	2or 72	. . 3 ((a ∩ (a ∩ b)) ∪ (a⊥ ∩ (a ∩ b)⊥ )) = ((a ∩ b) ∪ a⊥ )
12		ax-a2 31	. . 3 ((a ∩ b) ∪ a⊥ ) = (a⊥ ∪ (a ∩ b))
13	11, 12	ax-r2 36	. 2 ((a ∩ (a ∩ b)) ∪ (a⊥ ∩ (a ∩ b)⊥ )) = (a⊥ ∪ (a ∩ b))
14		dfb 94	. 2 (a ≡ (a ∩ b)) = ((a ∩ (a ∩ b)) ∪ (a⊥ ∩ (a ∩ b)⊥ ))
15		df-i1 44	. 2 (a →1 b) = (a⊥ ∪ (a ∩ b))
16	13, 14, 15	3tr1 63	1 (a ≡ (a ∩ b)) = (a →1 b)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7   →₁ wi1 12

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-b 39  df-a 40  df-t 41  df-f 42  df-i1 44

This theorem is used by:  nom35  324  nom55  336