I am currently trying to complete past papers but I don't understand a couple of questions on propositional and predicate calculi. I can only find very simple examples of normalisation/simplifying.
Normalise and where appropriate simplify the following expressions:
i) ¬(X ∧ Z) v (¬X v Y) ii) ¬Q v (¬Q ∧ R ∧ ¬T) v (R ∧ ¬Q ∧ T) iii) E v D → E
'Normalization' in this context typically means putting the statement into one of the 'Normal Forms', of which there are several:
'Boolean Statement' (this is not an official name and in fact not even an official 'Normal Form' (though I propose we could call it 'Boolean Normal Form'), but it is usually the first step into getting a statement into any of the other Normal Forms: A Boolean statement is one that uses only the 'Boolean operators' $\neg$, $\land$, and $\lor$. So, for example, the third statement contains an operator that is not boolean (though it is still 'truth-functional') the $\rightarrow$. Assuming this statement means $(E \lor D) \rightarrow E$ (you really need to add parentheses to disambiguate this statement!), we can use the equivalence $P \rightarrow Q \Leftrightarrow \neg P \lor Q$ to rewrite any $\rightarrow$ with Boolean operators. So in this case, we get $\neg(E \lor D) \lor E$
Negation Normal Form: A statement is in Negation Normal Form (NNF) when the only negations in the statement are negations of atomic statements (i.e. you cannot have a negation of some complex claim like a conjunction, disjunction, or negation itself). To put a statement into NNF, simply keep applying Demorgan's to work your negations 'inside' ... and remove any double negations. For example, consider what we just obtained for the third statement:
$\neg(E \lor D) \lor E$
The negation is a negation of a disjunction, so do a DeMorgan:
$(\neg E \land \neg D) \lor E$
And now it is in Negation Normal Form
'Disjunctive Normal Form': A statement is in DNF when it is a generalized disjunction of conjunctions of literals, where a literal is an atomic statement or a negation of an atomic statement.
'Conjunctive Normal Form': A statement is in CNF when it is a generalized conjunction of disjunctions of literals, where a literal is an atomic statement or a negation of an atomic statement.
(in other words, the CNF and DNF are completely 'symmetrical')
Since in a statement that is in NNF all negations are of atomic statements, it must be a recursive structure using any number of disjunctions and conjunctions, culminating with literals, so it is well on it way of being in DNF or CNF, but what you want to do now is 'flatten' this out, meaning that for a statement in DNF you want to use repeated Distribution of $\land$ over $\lor$: $P \land (Q \lor R) \Leftrightarrow (P \land Q) \lor (P \land R)$ until you have no more of such structures, and to get a statement into CNF you use repeated Distribution of $\lor$ over $\land$: $P \lor (Q \land R) \Leftrightarrow (P \lor Q) \land (P \lor R)$
Now, as it so happens, the statement we just got ($(\neg E \land \neg D) \lor E$) already is in DNF, since it is a disjunction of two disjuncts ($\neg E \land \neg D$ and $E$), and each of those disjuncts is a conjunction of literals (the first is a conjunction of two conjuncts $\neg E$ and $\neg D$, and the second is a conjunction of one conjunct $E$.
(yes, that last one feels a bit weird, since there is no explicit $\land$ being used, but a generalized conjunction conjuncts together a bunch of statements, and if that 'bunch' happens to be just one statement, then that one statement is the result on conjuncting 'all' of them together. Indeed, a statement like $E$ is in fact in DNF and CNF all by itself!)
$(\neg E \land \neg D) \lor E$ is not in CNF though, so we need to distribute the $\lor$ over the $\land$:
$(\neg E \lor E) \land (\neg D \lor E)$
And now it is in CNF.
Can you figure out the NNF, CNF, and DNF of the first two statements? (which are already in 'Boolean form')