How do I write the next equation?

$ A /cup B =(A-B) $

Lemma 4.3 and Theorem 4.5 provide a method to construct a proof for a given tautology in terms of the axioms of AXSP3B, as we show next with an example.
Example. The formula $ p \vee  (\neg (p \wedge \neg p)) $ is a tautology as shown by its true table. The formula is a disjunction, we put $ \eta = (\neg (p \wedge \neg p) $ . We prove that this formula is a Theorem by using Lemma 4.3.
Let $ \nu $ be a Valuation such that $ \nu (p)=0 $ , then $ \nu (\eta )=2 $. According to case 2 (e) of Lemma 4.3 $ \neg (p \wedge p) \vdash  \neg (p \wedge p) $ and $ \neg (p \wedge p) \vdash \neg (\neg \eta \wedge  \neg \eta )$
According to axiom PB10 $ \neg (\neg \eta \wedge \neg \eta )) \wedge \neg (p \wedge p) \rightarrow \neg (\neg (\eta \vee p) \wedge \neg (\eta \vee p)) $. According to axiom PB1
$ \neg (\neg (\eta \vee p) \wedge \neg (\eta \vee p)) \rightarrow (\eta \vee p $. Thus we obtain
$ \neg (p \wedge p) \vdash p \vee (\neg (p \wedge \neg p) $
Now let $ \nu $ be a valuation such that $ \nu (p)=1 $, then $ \nu (\eta )=0 $ and $ \nu (p \vee \eta )=1 $. According to Lemma 4.3 case 2 (c) $  (p \wedge \neg p) \vdash   (p \wedge \neg p) $ and $ (p \wedge \neg p) \vdash   \neg (\eta \wedge \eta )$. By axiom PB8 we have
$ ( \neg ( \eta \wedge  \eta )\wedge (p \wedge \neg p))\ rightarrow (\eta \vee p)\wedge \neg ((\eta \vee p)) $. And by POS 3 we have $ (\eta \vee p)\wedge \neg ((\eta \vee p))\rightarrow (\eta \vee p) $. So we obtain $ p\wedge \neg p  \vdash \neg (p \wedge \neg p) \vee p $

How do I write the next equation? \( A \cup B =(A-B) \)

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