Theorem u1lemnoa 660

Description: Lemma for Sasaki implication study. (Contributed by NM, 16-Dec-1997.)

Assertion

Ref	Expression
u1lemnoa	((a →1 b)⊥ ∪ a) = a

Proof of Theorem u1lemnoa

Step	Hyp	Ref	Expression
1		anor1 88	. . . 4 ((a →1 b) ∩ a⊥ ) = ((a →1 b)⊥ ∪ a)⊥
2	1	ax-r1 35	. . 3 ((a →1 b)⊥ ∪ a)⊥ = ((a →1 b) ∩ a⊥ )
3		u1lemana 605	. . 3 ((a →1 b) ∩ a⊥ ) = a⊥
4	2, 3	ax-r2 36	. 2 ((a →1 b)⊥ ∪ a)⊥ = a⊥
5	4	con1 66	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:  u1lem1  734