Theorem u2lemoa 603

Description: Lemma for Dishkant implication study.

Assertion

Ref	Expression
u2lemoa	((a →2 b) ∪ a) = 1

Proof of Theorem u2lemoa

Step	Hyp	Ref	Expression
1		df-i2 44	. . 3 (a →2 b) = (b ∪ (a⊥ ∩ b⊥ ))
2	1	ax-r5 37	. 2 ((a →2 b) ∪ a) = ((b ∪ (a⊥ ∩ b⊥ )) ∪ a)
3		ax-a2 30	. . 3 ((b ∪ (a⊥ ∩ b⊥ )) ∪ a) = (a ∪ (b ∪ (a⊥ ∩ b⊥ )))
4		ax-a3 31	. . . . 5 ((a ∪ b) ∪ (a⊥ ∩ b⊥ )) = (a ∪ (b ∪ (a⊥ ∩ b⊥ )))
5	4	ax-r1 34	. . . 4 (a ∪ (b ∪ (a⊥ ∩ b⊥ ))) = ((a ∪ b) ∪ (a⊥ ∩ b⊥ ))
6		ax-a2 30	. . . . 5 ((a ∪ b) ∪ (a⊥ ∩ b⊥ )) = ((a⊥ ∩ b⊥ ) ∪ (a ∪ b))
7		oran 79	. . . . . . 7 (a ∪ b) = (a⊥ ∩ b⊥ )⊥
8	7	lor 66	. . . . . 6 ((a⊥ ∩ b⊥ ) ∪ (a ∪ b)) = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ )⊥ )
9		df-t 40	. . . . . . 7 1 = ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ )⊥ )
10	9	ax-r1 34	. . . . . 6 ((a⊥ ∩ b⊥ ) ∪ (a⊥ ∩ b⊥ )⊥ ) = 1
11	8, 10	ax-r2 35	. . . . 5 ((a⊥ ∩ b⊥ ) ∪ (a ∪ b)) = 1
12	6, 11	ax-r2 35	. . . 4 ((a ∪ b) ∪ (a⊥ ∩ b⊥ )) = 1
13	5, 12	ax-r2 35	. . 3 (a ∪ (b ∪ (a⊥ ∩ b⊥ ))) = 1
14	3, 13	ax-r2 35	. 2 ((b ∪ (a⊥ ∩ b⊥ )) ∪ a) = 1
15	2, 14	ax-r2 35	1 ((a →2 b) ∪ a) = 1

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 9   →₂ wi2 14

This theorem is referenced by:  u2lemnana 628

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

This theorem depends on definitions:  df-a 39  df-t 40  df-i2 44

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