# Test

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? $$A \cup B =(A-B)$$