Theorem ska2b 219

Description: Axiom KA2b in Pavicic and Megill, 1998

Assertion

Ref	Expression
ska2b	(((a v c) == (b v c)) == ((a_|_ ^ c_|_)_|_ == (b_|_ ^ c_|_)_|_)) = 1

Proof of Theorem ska2b

Step	Hyp	Ref	Expression
1		oran 79	. . 3 (a v c) = (a_|_ ^ c_|_)_|_
2		oran 79	. . 3 (b v c) = (b_|_ ^ c_|_)_|_
3	1, 2	2bi 91	. 2 ((a v c) == (b v c)) = ((a_|_ ^ c_|_)_|_ == (b_|_ ^ c_|_)_|_)
4	3	bi1 110	1 (((a v c) == (b v c)) == ((a_|_ ^ c_|_)_|_ == (b_|_ ^ c_|_)_|_)) = 1

Syntax hints:   = wb 1  _|_wn 4   == tb 5   v wo 6   ^ wa 7  1wt 9

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

This theorem depends on definitions:  df-b 38  df-a 39  df-t 40  df-f 41

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