Theorem u1lem5 761

Description: Lemma for unified implication study. (Contributed by NM, 20-Dec-1997.)

Assertion

Ref	Expression
u1lem5	(a →1 (a →1 b)) = (a →1 b)

Proof of Theorem u1lem5

Step	Hyp	Ref	Expression
1		df-i1 44	. 2 (a →1 (a →1 b)) = (a⊥ ∪ (a ∩ (a →1 b)))
2		ancom 74	. . . . 5 (a ∩ (a →1 b)) = ((a →1 b) ∩ a)
3		u1lemaa 600	. . . . 5 ((a →1 b) ∩ a) = (a ∩ b)
4	2, 3	ax-r2 36	. . . 4 (a ∩ (a →1 b)) = (a ∩ b)
5	4	lor 70	. . 3 (a⊥ ∪ (a ∩ (a →1 b))) = (a⊥ ∪ (a ∩ b))
6		df-i1 44	. . . 4 (a →1 b) = (a⊥ ∪ (a ∩ b))
7	6	ax-r1 35	. . 3 (a⊥ ∪ (a ∩ b)) = (a →1 b)
8	5, 7	ax-r2 36	. 2 (a⊥ ∪ (a ∩ (a →1 b))) = (a →1 b)
9	1, 8	ax-r2 36	1 (a →1 (a →1 b)) = (a →1 b)

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

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-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by: (None)