Theorem u2lemc4 702

Description: Lemma for Dishkant implication study. (Contributed by NM, 24-Dec-1997.)

Hypothesis

Ref	Expression
ulemc3.1	a C b

Assertion

Ref	Expression
u2lemc4	(a →2 b) = (a⊥ ∪ b)

Proof of Theorem u2lemc4

Step	Hyp	Ref	Expression
1		df-i2 45	. 2 (a →2 b) = (b ∪ (a⊥ ∩ b⊥ ))
2		ulemc3.1	. . . . 5 a C b
3	2	comcom3 454	. . . 4 a⊥ C b
4	2	comcom4 455	. . . 4 a⊥ C b⊥
5	3, 4	fh4 472	. . 3 (b ∪ (a⊥ ∩ b⊥ )) = ((b ∪ a⊥ ) ∩ (b ∪ b⊥ ))
6		ax-a2 31	. . . . 5 (b ∪ a⊥ ) = (a⊥ ∪ b)
7		df-t 41	. . . . . 6 1 = (b ∪ b⊥ )
8	7	ax-r1 35	. . . . 5 (b ∪ b⊥ ) = 1
9	6, 8	2an 79	. . . 4 ((b ∪ a⊥ ) ∩ (b ∪ b⊥ )) = ((a⊥ ∪ b) ∩ 1)
10		an1 106	. . . 4 ((a⊥ ∪ b) ∩ 1) = (a⊥ ∪ b)
11	9, 10	ax-r2 36	. . 3 ((b ∪ a⊥ ) ∩ (b ∪ b⊥ )) = (a⊥ ∪ b)
12	5, 11	ax-r2 36	. 2 (b ∪ (a⊥ ∩ b⊥ )) = (a⊥ ∪ b)
13	1, 12	ax-r2 36	1 (a →2 b) = (a⊥ ∪ b)

This proof depends on syntax axioms:   = wb 1   C wc 3  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →₂ wi2 13

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-i2 45  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by:  u2lemle1  711