Theorem u3lem11a 787

Description: Lemma for unified implication study. (Contributed by NM, 18-Jan-1998.)

Assertion

Ref	Expression
u3lem11a	(a →3 ((b →3 a) →3 (a →3 b))⊥ ) = (a →3 b⊥ )

Proof of Theorem u3lem11a

Step	Hyp	Ref	Expression
1		ud3lem1 570	. . . . 5 ((b →3 a) →3 (a →3 b)) = (b ∪ (b⊥ ∩ a⊥ ))
2		ancom 74	. . . . . . . 8 (b⊥ ∩ a⊥ ) = (a⊥ ∩ b⊥ )
3		anor3 90	. . . . . . . 8 (a⊥ ∩ b⊥ ) = (a ∪ b)⊥
4	2, 3	ax-r2 36	. . . . . . 7 (b⊥ ∩ a⊥ ) = (a ∪ b)⊥
5	4	lor 70	. . . . . 6 (b ∪ (b⊥ ∩ a⊥ )) = (b ∪ (a ∪ b)⊥ )
6		oran1 91	. . . . . 6 (b ∪ (a ∪ b)⊥ ) = (b⊥ ∩ (a ∪ b))⊥
7	5, 6	ax-r2 36	. . . . 5 (b ∪ (b⊥ ∩ a⊥ )) = (b⊥ ∩ (a ∪ b))⊥
8	1, 7	ax-r2 36	. . . 4 ((b →3 a) →3 (a →3 b)) = (b⊥ ∩ (a ∪ b))⊥
9	8	con2 67	. . 3 ((b →3 a) →3 (a →3 b))⊥ = (b⊥ ∩ (a ∪ b))
10	9	ud3lem0a 260	. 2 (a →3 ((b →3 a) →3 (a →3 b))⊥ ) = (a →3 (b⊥ ∩ (a ∪ b)))
11		u3lem11 786	. 2 (a →3 (b⊥ ∩ (a ∪ b))) = (a →3 b⊥ )
12	10, 11	ax-r2 36	1 (a →3 ((b →3 a) →3 (a →3 b))⊥ ) = (a →3 b⊥ )

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₃ wi3 14

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-i3 46  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by: (None)