Theorem ska2b 227

Description: Axiom KA2b in Pavicic and Megill, 1998 (Contributed by NM, 9-Nov-1998.)

Assertion

Ref	Expression
ska2b	(((a ∪ c) ≡ (b ∪ c)) ≡ ((a⊥ ∩ c⊥ )⊥ ≡ (b⊥ ∩ c⊥ )⊥ )) = 1

Proof of Theorem ska2b

Step	Hyp	Ref	Expression
1		oran 87	. . 3 (a ∪ c) = (a⊥ ∩ c⊥ )⊥
2		oran 87	. . 3 (b ∪ c) = (b⊥ ∩ c⊥ )⊥
3	1, 2	2bi 99	. 2 ((a ∪ c) ≡ (b ∪ c)) = ((a⊥ ∩ c⊥ )⊥ ≡ (b⊥ ∩ c⊥ )⊥ )
4	3	bi1 118	1 (((a ∪ c) ≡ (b ∪ c)) ≡ ((a⊥ ∩ c⊥ )⊥ ≡ (b⊥ ∩ c⊥ )⊥ )) = 1

Syntax hints:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7  1wt 8

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42

This theorem is referenced by: (None)