Theorem lem4.6.6i3j0 1098

Description: Equation 4.14 of [MegPav2000] p. 23. The variable i in the paper is set to 3, and j is set to 0. (Contributed by Roy F. Longton, 1-Jul-2005.)

Assertion

Ref	Expression
lem4.6.6i3j0	((a →3 b) ∪ (a →0 b)) = (a →0 b)

Proof of Theorem lem4.6.6i3j0

Step	Hyp	Ref	Expression
1		ax-a3 32	. . 3 ((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ∪ (a⊥ ∪ b)) = (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ ((a ∩ (a⊥ ∪ b)) ∪ (a⊥ ∪ b)))
2		ax-a3 32	. . . . 5 (((a ∩ (a⊥ ∪ b)) ∪ a⊥ ) ∪ b) = ((a ∩ (a⊥ ∪ b)) ∪ (a⊥ ∪ b))
3	2	ax-r1 35	. . . 4 ((a ∩ (a⊥ ∪ b)) ∪ (a⊥ ∪ b)) = (((a ∩ (a⊥ ∪ b)) ∪ a⊥ ) ∪ b)
4	3	lor 70	. . 3 (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ ((a ∩ (a⊥ ∪ b)) ∪ (a⊥ ∪ b))) = (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (((a ∩ (a⊥ ∪ b)) ∪ a⊥ ) ∪ b))
5		ax-a2 31	. . . . . . 7 ((a ∩ (a⊥ ∪ b)) ∪ a⊥ ) = (a⊥ ∪ (a ∩ (a⊥ ∪ b)))
6		omln 446	. . . . . . 7 (a⊥ ∪ (a ∩ (a⊥ ∪ b))) = (a⊥ ∪ b)
7	5, 6	ax-r2 36	. . . . . 6 ((a ∩ (a⊥ ∪ b)) ∪ a⊥ ) = (a⊥ ∪ b)
8	7	ax-r5 38	. . . . 5 (((a ∩ (a⊥ ∪ b)) ∪ a⊥ ) ∪ b) = ((a⊥ ∪ b) ∪ b)
9	8	lor 70	. . . 4 (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (((a ∩ (a⊥ ∪ b)) ∪ a⊥ ) ∪ b)) = (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ ((a⊥ ∪ b) ∪ b))
10		leid 148	. . . . . . 7 (a⊥ ∪ b) ≤ (a⊥ ∪ b)
11		leor 159	. . . . . . 7 b ≤ (a⊥ ∪ b)
12	10, 11	lel2or 170	. . . . . 6 ((a⊥ ∪ b) ∪ b) ≤ (a⊥ ∪ b)
13		leo 158	. . . . . 6 (a⊥ ∪ b) ≤ ((a⊥ ∪ b) ∪ b)
14	12, 13	lebi 145	. . . . 5 ((a⊥ ∪ b) ∪ b) = (a⊥ ∪ b)
15	14	lor 70	. . . 4 (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ ((a⊥ ∪ b) ∪ b)) = (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a⊥ ∪ b))
16		leao1 162	. . . . . 6 (a⊥ ∩ b) ≤ (a⊥ ∪ b)
17		leao1 162	. . . . . 6 (a⊥ ∩ b⊥ ) ≤ (a⊥ ∪ b)
18	16, 17	lel2or 170	. . . . 5 ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ≤ (a⊥ ∪ b)
19	18	df-le2 131	. . . 4 (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a⊥ ∪ b)) = (a⊥ ∪ b)
20	9, 15, 19	3tr 65	. . 3 (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (((a ∩ (a⊥ ∪ b)) ∪ a⊥ ) ∪ b)) = (a⊥ ∪ b)
21	1, 4, 20	3tr 65	. 2 ((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ∪ (a⊥ ∪ b)) = (a⊥ ∪ b)
22		df-i3 46	. . 3 (a →3 b) = (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b)))
23		df-i0 43	. . 3 (a →0 b) = (a⊥ ∪ b)
24	22, 23	2or 72	. 2 ((a →3 b) ∪ (a →0 b)) = ((((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ∪ (a⊥ ∪ b))
25	21, 24, 23	3tr1 63	1 ((a →3 b) ∪ (a →0 b)) = (a →0 b)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₀ wi0 11   →₃ wi3 14

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  ax-r3 439

This proof depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i0 43  df-i3 46  df-le1 130  df-le2 131

This theorem is used by: (None)