Theorem oa6to4h1 935

Description: Satisfaction of 6-variable OA law hypothesis.

Hypotheses

Ref	Expression
oa6to4.1	b⊥ = (a →1 g)⊥
oa6to4.2	d⊥ = (c →1 g)⊥
oa6to4.3	f⊥ = (e →1 g)⊥

Assertion

Ref	Expression
oa6to4h1	a⊥ ≤ b⊥ ⊥

Proof of Theorem oa6to4h1

Step	Hyp	Ref	Expression
1		leo 150	. 2 a⊥ ≤ (a⊥ ∪ (a ∩ g))
2		oa6to4.1	. . . . 5 b⊥ = (a →1 g)⊥
3		df-i1 43	. . . . . 6 (a →1 g) = (a⊥ ∪ (a ∩ g))
4	3	ax-r4 36	. . . . 5 (a →1 g)⊥ = (a⊥ ∪ (a ∩ g))⊥
5	2, 4	ax-r2 35	. . . 4 b⊥ = (a⊥ ∪ (a ∩ g))⊥
6	5	ax-r1 34	. . 3 (a⊥ ∪ (a ∩ g))⊥ = b⊥
7	6	con3 65	. 2 (a⊥ ∪ (a ∩ g)) = b⊥ ⊥
8	1, 7	lbtr 131	1 a⊥ ≤ b⊥ ⊥

Syntax hints:   = wb 1   ≤ wle 2  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 13

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37

This theorem depends on definitions:  df-a 39  df-i1 43  df-le1 122  df-le2 123

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