Theorem u1lemnanb 655

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

Assertion

Ref	Expression
u1lemnanb	((a →1 b)⊥ ∩ b⊥ ) = (a ∩ b⊥ )

Proof of Theorem u1lemnanb

Step	Hyp	Ref	Expression
1		u1lemob 630	. . 3 ((a →1 b) ∪ b) = (a⊥ ∪ b)
2		oran 87	. . 3 ((a →1 b) ∪ b) = ((a →1 b)⊥ ∩ b⊥ )⊥
3		oran2 92	. . 3 (a⊥ ∪ b) = (a ∩ b⊥ )⊥
4	1, 2, 3	3tr2 64	. 2 ((a →1 b)⊥ ∩ b⊥ )⊥ = (a ∩ b⊥ )⊥
5	4	con1 66	1 ((a →1 b)⊥ ∩ b⊥ ) = (a ∩ b⊥ )

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12

This theorem was proved from 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 theorem depends on definitions:  df-a 40  df-i1 44  df-le1 130  df-le2 131

This theorem is referenced by:  u3lem14a  791