Theorem ska2a 226

Description: Axiom KA2a in Pavicic and Megill, 1998 (Contributed by NM, 9-Nov-1998.)

Assertion

Ref	Expression
ska2a	(((a ∪ c) ≡ (b ∪ c)) ≡ ((c ∪ a) ≡ (c ∪ b))) = 1

Proof of Theorem ska2a

Step	Hyp	Ref	Expression
1		ax-a2 31	. . 3 (a ∪ c) = (c ∪ a)
2		ax-a2 31	. . 3 (b ∪ c) = (c ∪ b)
3	1, 2	2bi 99	. 2 ((a ∪ c) ≡ (b ∪ c)) = ((c ∪ a) ≡ (c ∪ b))
4	3	bi1 118	1 (((a ∪ c) ≡ (b ∪ c)) ≡ ((c ∪ a) ≡ (c ∪ b))) = 1

Syntax hints:   = wb 1   ≡ tb 5   ∪ wo 6  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)