Theorem u1lem7 772

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

Assertion

Ref	Expression
u1lem7	(a →1 (a⊥ →1 b)) = 1

Proof of Theorem u1lem7

Step	Hyp	Ref	Expression
1		df-i1 44	. 2 (a →1 (a⊥ →1 b)) = (a⊥ ∪ (a ∩ (a⊥ →1 b)))
2		ax-a1 30	. . . . . 6 a = a⊥ ⊥
3	2	ran 78	. . . . 5 (a ∩ (a⊥ →1 b)) = (a⊥ ⊥ ∩ (a⊥ →1 b))
4		ancom 74	. . . . . 6 (a⊥ ⊥ ∩ (a⊥ →1 b)) = ((a⊥ →1 b) ∩ a⊥ ⊥ )
5		u1lemana 605	. . . . . 6 ((a⊥ →1 b) ∩ a⊥ ⊥ ) = a⊥ ⊥
6	4, 5	ax-r2 36	. . . . 5 (a⊥ ⊥ ∩ (a⊥ →1 b)) = a⊥ ⊥
7	3, 6	ax-r2 36	. . . 4 (a ∩ (a⊥ →1 b)) = a⊥ ⊥
8	7	lor 70	. . 3 (a⊥ ∪ (a ∩ (a⊥ →1 b))) = (a⊥ ∪ a⊥ ⊥ )
9		df-t 41	. . . 4 1 = (a⊥ ∪ a⊥ ⊥ )
10	9	ax-r1 35	. . 3 (a⊥ ∪ a⊥ ⊥ ) = 1
11	8, 10	ax-r2 36	. 2 (a⊥ ∪ (a ∩ (a⊥ →1 b))) = 1
12	1, 11	ax-r2 36	1 (a →1 (a⊥ →1 b)) = 1

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →₁ wi1 12

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This proof depends on definitions:  df-a 40  df-t 41  df-i1 44

This theorem is used by: (None)