Theorem u2lemnonb 676

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

Assertion

Ref	Expression
u2lemnonb	((a →2 b)⊥ ∪ b⊥ ) = b⊥

Proof of Theorem u2lemnonb

Step	Hyp	Ref	Expression
1		df-a 40	. . . 4 ((a →2 b) ∩ b) = ((a →2 b)⊥ ∪ b⊥ )⊥
2	1	ax-r1 35	. . 3 ((a →2 b)⊥ ∪ b⊥ )⊥ = ((a →2 b) ∩ b)
3		u2lemab 611	. . 3 ((a →2 b) ∩ b) = b
4	2, 3	ax-r2 36	. 2 ((a →2 b)⊥ ∪ b⊥ )⊥ = b
5	4	con3 68	1 ((a →2 b)⊥ ∪ b⊥ ) = b⊥

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

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-i2 45

This theorem is used by: (None)