Theorem u1lemnanb 637

Description: Lemma for Sasaki implication study.

Assertion

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

Proof of Theorem u1lemnanb

Step	Hyp	Ref	Expression
1		u1lemob 612	. . 3 ((a →1 b) ∪ b) = (a⊥ ∪ b)
2		oran 79	. . 3 ((a →1 b) ∪ b) = ((a →1 b)⊥ ∩ b⊥ )⊥
3		oran2 84	. . 3 (a⊥ ∪ b) = (a ∩ b⊥ )⊥
4	1, 2, 3	3tr2 61	. 2 ((a →1 b)⊥ ∩ b⊥ )⊥ = (a ∩ b⊥ )⊥
5	4	con1 63	1 ((a →1 b)⊥ ∩ b⊥ ) = (a ∩ b⊥ )

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 13

This theorem is referenced by:  u3lem14a 773

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37

This theorem depends on definitions:  df-a 39  df-i1 43  df-le1 122  df-le2 123

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