Theorem ud3lem3a 554

Description: Lemma for unified disjunction.

Assertion

Ref	Expression
ud3lem3a	((a →3 b)⊥ ∩ (a ∪ b)) = (a →3 b)⊥

Proof of Theorem ud3lem3a

Step	Hyp	Ref	Expression
1		ud3lem0c 271	. . 3 (a →3 b)⊥ = (((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a⊥ ∪ (a ∩ b⊥ )))
2		lea 152	. . . 4 (((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a⊥ ∪ (a ∩ b⊥ ))) ≤ ((a ∪ b⊥ ) ∩ (a ∪ b))
3		lear 153	. . . 4 ((a ∪ b⊥ ) ∩ (a ∪ b)) ≤ (a ∪ b)
4	2, 3	letr 129	. . 3 (((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a⊥ ∪ (a ∩ b⊥ ))) ≤ (a ∪ b)
5	1, 4	bltr 130	. 2 (a →3 b)⊥ ≤ (a ∪ b)
6	5	df2le2 128	1 ((a →3 b)⊥ ∩ (a ∪ b)) = (a →3 b)⊥

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₃ wi3 15

This theorem is referenced by:  ud3lem3 558

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

This theorem depends on definitions:  df-a 39  df-i3 45  df-le1 122  df-le2 123

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