Theorem oa4to6dual 944

Description: Lemma for orthoarguesian law (4-variable to 6-variable proof).

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 943	. 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 129	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 13

This theorem is referenced by:  oa4to6 945  oa3-6to3 967  oa3-2to4 968  oa3-u1 971  oa3-u2 972

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a4 32  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37  ax-r3 421

This theorem depends on definitions:  df-b 38  df-a 39  df-t 40  df-f 41  df-i1 43  df-le1 122  df-le2 123  df-c1 124  df-c2 125

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