Theorem u4lem3n 755

Description: Lemma for unified implication study. (Contributed by NM, 17-Dec-1997.)

Assertion

Ref	Expression
u4lem3n	(a →4 (b →4 a))⊥ = (a ∩ ((a⊥ ∪ b) ∩ (a⊥ ∪ b⊥ )))

Proof of Theorem u4lem3n

Step	Hyp	Ref	Expression
1		u4lem3 752	. . 3 (a →4 (b →4 a)) = (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))
2		ax-a2 31	. . . . . 6 ((a ∩ b) ∪ (a ∩ b⊥ )) = ((a ∩ b⊥ ) ∪ (a ∩ b))
3		anor1 88	. . . . . . . 8 (a ∩ b⊥ ) = (a⊥ ∪ b)⊥
4		df-a 40	. . . . . . . 8 (a ∩ b) = (a⊥ ∪ b⊥ )⊥
5	3, 4	2or 72	. . . . . . 7 ((a ∩ b⊥ ) ∪ (a ∩ b)) = ((a⊥ ∪ b)⊥ ∪ (a⊥ ∪ b⊥ )⊥ )
6		oran3 93	. . . . . . 7 ((a⊥ ∪ b)⊥ ∪ (a⊥ ∪ b⊥ )⊥ ) = ((a⊥ ∪ b) ∩ (a⊥ ∪ b⊥ ))⊥
7	5, 6	ax-r2 36	. . . . . 6 ((a ∩ b⊥ ) ∪ (a ∩ b)) = ((a⊥ ∪ b) ∩ (a⊥ ∪ b⊥ ))⊥
8	2, 7	ax-r2 36	. . . . 5 ((a ∩ b) ∪ (a ∩ b⊥ )) = ((a⊥ ∪ b) ∩ (a⊥ ∪ b⊥ ))⊥
9	8	lor 70	. . . 4 (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ ))) = (a⊥ ∪ ((a⊥ ∪ b) ∩ (a⊥ ∪ b⊥ ))⊥ )
10		oran3 93	. . . 4 (a⊥ ∪ ((a⊥ ∪ b) ∩ (a⊥ ∪ b⊥ ))⊥ ) = (a ∩ ((a⊥ ∪ b) ∩ (a⊥ ∪ b⊥ )))⊥
11	9, 10	ax-r2 36	. . 3 (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ ))) = (a ∩ ((a⊥ ∪ b) ∩ (a⊥ ∪ b⊥ )))⊥
12	1, 11	ax-r2 36	. 2 (a →4 (b →4 a)) = (a ∩ ((a⊥ ∪ b) ∩ (a⊥ ∪ b⊥ )))⊥
13	12	con2 67	1 (a →4 (b →4 a))⊥ = (a ∩ ((a⊥ ∪ b) ∩ (a⊥ ∪ b⊥ )))

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₄ wi4 15

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-r3 439

This proof depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i4 47  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by: (None)