Theorem oa6to4h3 957

Description: Satisfaction of 6-variable OA law hypothesis. (Contributed by NM, 22-Dec-1998.)

Hypotheses

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

Assertion

Ref	Expression
oa6to4h3	e⊥ ≤ f⊥ ⊥

Proof of Theorem oa6to4h3

Step	Hyp	Ref	Expression
1		leo 158	. 2 e⊥ ≤ (e⊥ ∪ (e ∩ g))
2		oa6to4.3	. . . . 5 f⊥ = (e →1 g)⊥
3		df-i1 44	. . . . . 6 (e →1 g) = (e⊥ ∪ (e ∩ g))
4	3	ax-r4 37	. . . . 5 (e →1 g)⊥ = (e⊥ ∪ (e ∩ g))⊥
5	2, 4	ax-r2 36	. . . 4 f⊥ = (e⊥ ∪ (e ∩ g))⊥
6	5	ax-r1 35	. . 3 (e⊥ ∪ (e ∩ g))⊥ = f⊥
7	6	con3 68	. 2 (e⊥ ∪ (e ∩ g)) = f⊥ ⊥
8	1, 7	lbtr 139	1 e⊥ ≤ f⊥ ⊥

This proof depends on syntax axioms:   = wb 1   ≤ wle 2  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This proof depends on definitions:  df-a 40  df-i1 44  df-le1 130  df-le2 131

This theorem is used by: (None)