Theorem u1lemnab 650

Description: Lemma for Sasaki implication study.

Assertion

Ref	Expression
u1lemnab	((a →1 b)⊥ ∩ b) = 0

Proof of Theorem u1lemnab

Step	Hyp	Ref	Expression
1		u1lemonb 635	. . 3 ((a →1 b) ∪ b⊥ ) = 1
2		oran1 91	. . 3 ((a →1 b) ∪ b⊥ ) = ((a →1 b)⊥ ∩ b)⊥
3		df-f 42	. . . . 5 0 = 1⊥
4	3	con2 67	. . . 4 0⊥ = 1
5	4	ax-r1 35	. . 3 1 = 0⊥
6	1, 2, 5	3tr2 64	. 2 ((a →1 b)⊥ ∩ b)⊥ = 0⊥
7	6	con1 66	1 ((a →1 b)⊥ ∩ b) = 0

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8  0wf 9   →₁ wi1 12

This theorem is referenced by:  u1lemn1b 730

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This theorem depends on definitions:  df-a 40  df-t 41  df-f 42  df-i1 44

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