Theorem lem4.6.6i0j1 1088

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

Assertion

Ref	Expression
lem4.6.6i0j1	((a →0 b) ∪ (a →1 b)) = (a →0 b)

Proof of Theorem lem4.6.6i0j1

Step	Hyp	Ref	Expression
1		leid 148	. . . 4 (a⊥ ∪ b) ≤ (a⊥ ∪ b)
2		lear 161	. . . . 5 (a ∩ b) ≤ b
3	2	lelor 166	. . . 4 (a⊥ ∪ (a ∩ b)) ≤ (a⊥ ∪ b)
4	1, 3	lel2or 170	. . 3 ((a⊥ ∪ b) ∪ (a⊥ ∪ (a ∩ b))) ≤ (a⊥ ∪ b)
5		leo 158	. . 3 (a⊥ ∪ b) ≤ ((a⊥ ∪ b) ∪ (a⊥ ∪ (a ∩ b)))
6	4, 5	lebi 145	. 2 ((a⊥ ∪ b) ∪ (a⊥ ∪ (a ∩ b))) = (a⊥ ∪ b)
7		df-i0 43	. . 3 (a →0 b) = (a⊥ ∪ b)
8		df-i1 44	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
9	7, 8	2or 72	. 2 ((a →0 b) ∪ (a →1 b)) = ((a⊥ ∪ b) ∪ (a⊥ ∪ (a ∩ b)))
10	6, 9, 7	3tr1 63	1 ((a →0 b) ∪ (a →1 b)) = (a →0 b)

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₀ wi0 11   →₁ wi1 12

This theorem was proved from 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 theorem depends on definitions:  df-a 40  df-t 41  df-f 42  df-i0 43  df-i1 44  df-le1 130  df-le2 131

This theorem is referenced by: (None)