Theorem wql2lem4 291

Description: Lemma for →₂ WQL axiom. (Contributed by NM, 6-Dec-1998.)

Hypotheses

Ref	Expression
wql2lem4.1	(((a ∩ b⊥ ) ∪ (a ∩ b)) →2 (a⊥ ∪ (a ∩ b))) = 1
wql2lem4.2	((a →1 b) ∪ (a ∩ b⊥ )) = 1

Assertion

Ref	Expression
wql2lem4	(a →1 b) = 1

Proof of Theorem wql2lem4

Step	Hyp	Ref	Expression
1		df-i1 44	. 2 (a →1 b) = (a⊥ ∪ (a ∩ b))
2		id 59	. 2 (a⊥ ∪ (a ∩ b)) = (a⊥ ∪ (a ∩ b))
3		ax-a2 31	. . . 4 ((a ∩ b⊥ ) ∪ (a⊥ ∪ (a ∩ b))) = ((a⊥ ∪ (a ∩ b)) ∪ (a ∩ b⊥ ))
4	1	ax-r5 38	. . . . 5 ((a →1 b) ∪ (a ∩ b⊥ )) = ((a⊥ ∪ (a ∩ b)) ∪ (a ∩ b⊥ ))
5	4	ax-r1 35	. . . 4 ((a⊥ ∪ (a ∩ b)) ∪ (a ∩ b⊥ )) = ((a →1 b) ∪ (a ∩ b⊥ ))
6		wql2lem4.2	. . . 4 ((a →1 b) ∪ (a ∩ b⊥ )) = 1
7	3, 5, 6	3tr 65	. . 3 ((a ∩ b⊥ ) ∪ (a⊥ ∪ (a ∩ b))) = 1
8		wql2lem4.1	. . . 4 (((a ∩ b⊥ ) ∪ (a ∩ b)) →2 (a⊥ ∪ (a ∩ b))) = 1
9	8	wql2lem2 289	. . 3 (((a ∩ b⊥ ) ∪ (a⊥ ∪ (a ∩ b)))⊥ ∪ (a⊥ ∪ (a ∩ b))) = 1
10	7, 9	skr0 242	. 2 (a⊥ ∪ (a ∩ b)) = 1
11	1, 2, 10	3tr 65	1 (a →1 b) = 1

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →₁ wi1 12   →₂ wi2 13

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  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-f 42  df-i1 44  df-i2 45

This theorem is used by: (None)