Theorem oa4to6dual 964

Description: Lemma for orthoarguesian law (4-variable to 6-variable proof). (Contributed by NM, 19-Dec-1998.)

Hypotheses

Ref	Expression
oa4to6lem.1	a⊥ ≤ b
oa4to6lem.2	c⊥ ≤ d
oa4to6lem.3	e⊥ ≤ f
oa4to6lem.4	g = (((a ∩ b) ∪ (c ∩ d)) ∪ (e ∩ f))
oa4to6lem.oa4	((a →1 g) ∩ (a ∪ (c ∩ (((a ∩ c) ∪ ((a →1 g) ∩ (c →1 g))) ∪ (((a ∩ e) ∪ ((a →1 g) ∩ (e →1 g))) ∩ ((c ∩ e) ∪ ((c →1 g) ∩ (e →1 g)))))))) ≤ g

Assertion

Ref	Expression
oa4to6dual	(b ∩ (a ∪ (c ∩ (((a ∩ c) ∪ (b ∩ d)) ∪ (((a ∩ e) ∪ (b ∩ f)) ∩ ((c ∩ e) ∪ (d ∩ f))))))) ≤ g

Proof of Theorem oa4to6dual

Step	Hyp	Ref	Expression
1		oa4to6lem.1	. . 3 a⊥ ≤ b
2		oa4to6lem.2	. . 3 c⊥ ≤ d
3		oa4to6lem.3	. . 3 e⊥ ≤ f
4		oa4to6lem.4	. . 3 g = (((a ∩ b) ∪ (c ∩ d)) ∪ (e ∩ f))
5	1, 2, 3, 4	oa4to6lem4 963	. 2 (b ∩ (a ∪ (c ∩ (((a ∩ c) ∪ (b ∩ d)) ∪ (((a ∩ e) ∪ (b ∩ f)) ∩ ((c ∩ e) ∪ (d ∩ f))))))) ≤ ((a →1 g) ∩ (a ∪ (c ∩ (((a ∩ c) ∪ ((a →1 g) ∩ (c →1 g))) ∪ (((a ∩ e) ∪ ((a →1 g) ∩ (e →1 g))) ∩ ((c ∩ e) ∪ ((c →1 g) ∩ (e →1 g))))))))
6		oa4to6lem.oa4	. 2 ((a →1 g) ∩ (a ∪ (c ∩ (((a ∩ c) ∪ ((a →1 g) ∩ (c →1 g))) ∪ (((a ∩ e) ∪ ((a →1 g) ∩ (e →1 g))) ∩ ((c ∩ e) ∪ ((c →1 g) ∩ (e →1 g)))))))) ≤ g
7	5, 6	letr 137	1 (b ∩ (a ∪ (c ∩ (((a ∩ c) ∪ (b ∩ d)) ∪ (((a ∩ e) ∪ (b ∩ f)) ∩ ((c ∩ e) ∪ (d ∩ f))))))) ≤ g

Syntax hints:   = wb 1   ≤ wle 2  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-r3 439

This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is referenced by:  oa4to6  965  oa3-6to3  987  oa3-2to4  988  oa3-u1  991  oa3-u2  992