Theorem lem4.6.6i4j0 1100

Description: Equation 4.14 of [MegPav2000] p. 23. The variable i in the paper is set to 4, and j is set to 0. (Contributed by Roy F. Longton, 2-Jul-2005.)

Assertion

Ref	Expression
lem4.6.6i4j0	((a →4 b) ∪ (a →0 b)) = (a →0 b)

Proof of Theorem lem4.6.6i4j0

Step	Hyp	Ref	Expression
1		leao4 165	. . . . 5 (a ∩ b) ≤ (a⊥ ∪ b)
2		leao1 162	. . . . 5 (a⊥ ∩ b) ≤ (a⊥ ∪ b)
3	1, 2	lel2or 170	. . . 4 ((a ∩ b) ∪ (a⊥ ∩ b)) ≤ (a⊥ ∪ b)
4		lea 160	. . . 4 ((a⊥ ∪ b) ∩ b⊥ ) ≤ (a⊥ ∪ b)
5	3, 4	lel2or 170	. . 3 (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∪ b) ∩ b⊥ )) ≤ (a⊥ ∪ b)
6	5	df-le2 131	. 2 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∪ b) ∩ b⊥ )) ∪ (a⊥ ∪ b)) = (a⊥ ∪ b)
7		df-i4 47	. . 3 (a →4 b) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∪ b) ∩ b⊥ ))
8		df-i0 43	. . 3 (a →0 b) = (a⊥ ∪ b)
9	7, 8	2or 72	. 2 ((a →4 b) ∪ (a →0 b)) = ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∪ b) ∩ b⊥ )) ∪ (a⊥ ∪ b))
10	6, 9, 8	3tr1 63	1 ((a →4 b) ∪ (a →0 b)) = (a →0 b)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₀ wi0 11   →₄ wi4 15

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-i4 47  df-le1 130  df-le2 131

This theorem is used by: (None)