Theorem lem4.6.6i0j3 1090

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

Assertion

Ref	Expression
lem4.6.6i0j3	((a →0 b) ∪ (a →3 b)) = (a →0 b)

Proof of Theorem lem4.6.6i0j3

Step	Hyp	Ref	Expression
1		leid 148	. . . 4 (a⊥ ∪ b) ≤ (a⊥ ∪ b)
2		leao1 162	. . . . . 6 (a⊥ ∩ b) ≤ (a⊥ ∪ b)
3		leao1 162	. . . . . 6 (a⊥ ∩ b⊥ ) ≤ (a⊥ ∪ b)
4	2, 3	lel2or 170	. . . . 5 ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ≤ (a⊥ ∪ b)
5		lear 161	. . . . 5 (a ∩ (a⊥ ∪ b)) ≤ (a⊥ ∪ b)
6	4, 5	lel2or 170	. . . 4 (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))) ≤ (a⊥ ∪ b)
7	1, 6	lel2or 170	. . 3 ((a⊥ ∪ b) ∪ (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b)))) ≤ (a⊥ ∪ b)
8		leo 158	. . 3 (a⊥ ∪ b) ≤ ((a⊥ ∪ b) ∪ (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))))
9	7, 8	lebi 145	. 2 ((a⊥ ∪ b) ∪ (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b)))) = (a⊥ ∪ b)
10		df-i0 43	. . 3 (a →0 b) = (a⊥ ∪ b)
11		df-i3 46	. . 3 (a →3 b) = (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b)))
12	10, 11	2or 72	. 2 ((a →0 b) ∪ (a →3 b)) = ((a⊥ ∪ b) ∪ (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ (a ∩ (a⊥ ∪ b))))
13	9, 12, 10	3tr1 63	1 ((a →0 b) ∪ (a →3 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

This proof depends on definitions:  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)