Theorem u1lemnob 670

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

Assertion

Ref	Expression
u1lemnob	((a →1 b)⊥ ∪ b) = (a ∪ b)

Proof of Theorem u1lemnob

Step	Hyp	Ref	Expression
1		u1lemanb 615	. . 3 ((a →1 b) ∩ b⊥ ) = (a⊥ ∩ b⊥ )
2		anor1 88	. . 3 ((a →1 b) ∩ b⊥ ) = ((a →1 b)⊥ ∪ b)⊥
3		anor3 90	. . 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-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-r3 439

This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is referenced by: (None)