Theorem u3lem1n 723

Description: Lemma for unified implication study.

Assertion

Ref	Expression
u3lem1n	((a →3 b) →3 a)⊥ = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))

Proof of Theorem u3lem1n

Step	Hyp	Ref	Expression
1		u3lem1 718	. . 3 ((a →3 b) →3 a) = ((a ∪ b) ∩ (a ∪ b⊥ ))
2		ancom 68	. . . 4 ((a ∪ b) ∩ (a ∪ b⊥ )) = ((a ∪ b⊥ ) ∩ (a ∪ b))
3		df-a 39	. . . . 5 ((a ∪ b⊥ ) ∩ (a ∪ b)) = ((a ∪ b⊥ )⊥ ∪ (a ∪ b)⊥ )⊥
4		anor2 81	. . . . . . . 8 (a⊥ ∩ b) = (a ∪ b⊥ )⊥
5		anor3 82	. . . . . . . 8 (a⊥ ∩ b⊥ ) = (a ∪ b)⊥
6	4, 5	2or 67	. . . . . . 7 ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) = ((a ∪ b⊥ )⊥ ∪ (a ∪ b)⊥ )
7	6	ax-r4 36	. . . . . 6 ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥ = ((a ∪ b⊥ )⊥ ∪ (a ∪ b)⊥ )⊥
8	7	ax-r1 34	. . . . 5 ((a ∪ b⊥ )⊥ ∪ (a ∪ b)⊥ )⊥ = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥
9	3, 8	ax-r2 35	. . . 4 ((a ∪ b⊥ ) ∩ (a ∪ b)) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥
10	2, 9	ax-r2 35	. . 3 ((a ∪ b) ∩ (a ∪ b⊥ )) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥
11	1, 10	ax-r2 35	. 2 ((a →3 b) →3 a) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))⊥
12	11	con2 64	1 ((a →3 b) →3 a)⊥ = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))

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

This theorem is referenced by:  u3lem2 728

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-i3 45  df-le1 122  df-le2 123  df-c1 124  df-c2 125

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