Theorem u1lemonb 617

Description: Lemma for Sasaki implication study.

Assertion

Ref	Expression
u1lemonb	((a →1 b) ∪ b⊥ ) = 1

Proof of Theorem u1lemonb

Step	Hyp	Ref	Expression
1		df-i1 43	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
2	1	ax-r5 37	. 2 ((a →1 b) ∪ b⊥ ) = ((a⊥ ∪ (a ∩ b)) ∪ b⊥ )
3		or32 75	. . 3 ((a⊥ ∪ (a ∩ b)) ∪ b⊥ ) = ((a⊥ ∪ b⊥ ) ∪ (a ∩ b))
4		df-a 39	. . . . 5 (a ∩ b) = (a⊥ ∪ b⊥ )⊥
5	4	lor 66	. . . 4 ((a⊥ ∪ b⊥ ) ∪ (a ∩ b)) = ((a⊥ ∪ b⊥ ) ∪ (a⊥ ∪ b⊥ )⊥ )
6		df-t 40	. . . . 5 1 = ((a⊥ ∪ b⊥ ) ∪ (a⊥ ∪ b⊥ )⊥ )
7	6	ax-r1 34	. . . 4 ((a⊥ ∪ b⊥ ) ∪ (a⊥ ∪ b⊥ )⊥ ) = 1
8	5, 7	ax-r2 35	. . 3 ((a⊥ ∪ b⊥ ) ∪ (a ∩ b)) = 1
9	3, 8	ax-r2 35	. 2 ((a⊥ ∪ (a ∩ b)) ∪ b⊥ ) = 1
10	2, 9	ax-r2 35	1 ((a →1 b) ∪ b⊥ ) = 1

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 9   →₁ wi1 13

This theorem is referenced by:  u1lemnab 632  u3lem14a 773

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

This theorem depends on definitions:  df-a 39  df-t 40  df-i1 43

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