Theorem 1i1 266

Description: Antecedent of 1 on Sasaki conditional.

Assertion

Ref	Expression
1i1	(1 →1 a) = a

Proof of Theorem 1i1

Step	Hyp	Ref	Expression
1		df-i1 43	. 2 (1 →1 a) = (1⊥ ∪ (1 ∩ a))
2		df-f 41	. . . . 5 0 = 1⊥
3	2	ax-r1 34	. . . 4 1⊥ = 0
4		ancom 68	. . . . 5 (1 ∩ a) = (a ∩ 1)
5		an1 98	. . . . 5 (a ∩ 1) = a
6	4, 5	ax-r2 35	. . . 4 (1 ∩ a) = a
7	3, 6	2or 67	. . 3 (1⊥ ∪ (1 ∩ a)) = (0 ∪ a)
8		ax-a2 30	. . . 4 (0 ∪ a) = (a ∪ 0)
9		or0 94	. . . 4 (a ∪ 0) = a
10	8, 9	ax-r2 35	. . 3 (0 ∪ a) = a
11	7, 10	ax-r2 35	. 2 (1⊥ ∪ (1 ∩ a)) = a
12	1, 11	ax-r2 35	1 (1 →1 a) = a

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 9  0wf 10   →₁ wi1 13

This theorem is referenced by:  oa3-6lem 960

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

This theorem depends on definitions:  df-a 39  df-t 40  df-f 41  df-i1 43

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