Theorem u1lemana 605

Description: Lemma for Sasaki implication study. (Contributed by NM, 14-Dec-1997.)

Assertion

Ref	Expression
u1lemana	((a →1 b) ∩ a⊥ ) = a⊥

Proof of Theorem u1lemana

Step	Hyp	Ref	Expression
1		df-i1 44	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
2	1	ran 78	. 2 ((a →1 b) ∩ a⊥ ) = ((a⊥ ∪ (a ∩ b)) ∩ a⊥ )
3		ancom 74	. . 3 ((a⊥ ∪ (a ∩ b)) ∩ a⊥ ) = (a⊥ ∩ (a⊥ ∪ (a ∩ b)))
4		anabs 121	. . 3 (a⊥ ∩ (a⊥ ∪ (a ∩ b))) = a⊥
5	3, 4	ax-r2 36	. 2 ((a⊥ ∪ (a ∩ b)) ∩ a⊥ ) = a⊥
6	2, 5	ax-r2 36	1 ((a →1 b) ∩ a⊥ ) = a⊥

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-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

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

This theorem is used by:  u1lemnoa  660  u12lembi  726  u1lem7  772