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 $