Theorem lem4.6.6i1j0 1092

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

Assertion

Ref	Expression
lem4.6.6i1j0	((a →1 b) ∪ (a →0 b)) = (a →0 b)

Proof of Theorem lem4.6.6i1j0

Step	Hyp	Ref	Expression
1		lear 161	. . . 4 (a ∩ b) ≤ b
2	1	lelor 166	. . 3 (a⊥ ∪ (a ∩ b)) ≤ (a⊥ ∪ b)
3	2	df-le2 131	. 2 ((a⊥ ∪ (a ∩ b)) ∪ (a⊥ ∪ b)) = (a⊥ ∪ b)
4		df-i1 44	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
5		df-i0 43	. . 3 (a →0 b) = (a⊥ ∪ b)
6	4, 5	2or 72	. 2 ((a →1 b) ∪ (a →0 b)) = ((a⊥ ∪ (a ∩ b)) ∪ (a⊥ ∪ b))
7	3, 6, 5	3tr1 63	1 ((a →1 b) ∪ (a →0 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)