# Explanation on notation (propositional calculus)

I'm reading the book Mathematical logic by Cori and Lascar and, though I understand now what is a canonical disjunctive normal form, I can't understand what is written here:

Let $$X$$ be a non-empty subset of $$\{0,1\}^n$$ and let $$F_X$$ be the formula: $$$$\bigvee_{(\varepsilon_1,...,\varepsilon_n) \in X }\left(\bigwedge_{1\leq i \leq n}\varepsilon_i A_i\right)\$$$$

Then the formula $$F_X$$ is satisfied by those distributions of truth values $$\delta_{\varepsilon_1 ... \varepsilon_n}$$ for which $$(\varepsilon_1, ..., \varepsilon_n) \in X$$ and only by these.

We can read the formula inside the parentheses as a sort of algebraic multiplication : $$0A_1 ∧ 1A_2 ∧ \ldots ∧ 0A_n$$, that corresponds to formula :

$$¬A_1 ∧ A_2 ∧ \ldots ∧ ¬A_n$$.

See page 34 :

an element $$(\varepsilon_1, \varepsilon_2,\ldots, \varepsilon_n)$$ of $$\{ 0,1 \}^n$$ is an $$n$$-uple of booleans (aka : truth values).

And $$\delta_{\varepsilon_1, \varepsilon_2,\ldots, \varepsilon_n}$$ is the distribution of truth values that assigns to propositional variable $$A_i$$ the truth value $$\varepsilon_i$$.

For each propositional variable $$A$$ and for each $$\varepsilon \in \{ 0,1 \}$$ we have that :

$$\varepsilon A$$ denotes $$A$$ if $$\varepsilon =1$$ and $$\lnot A$$ if $$\varepsilon =0$$.

• Thanks. But that's not where i'm stuck: $\varepsilon_i A_i$ can only be defined in $X$ which can have less elements than $\{0,1\}^n$, so I don't get how we can intersect from $i$ to $n$. Dec 10, 2019 at 11:18
• @ZeD --- not clear... Presumably we have a formula $F$ which is a truth-function of $n$ prop variables $A_1,\ldots, A_n$. Thus, we need a corresponding $(\varepsilon_1,\ldots, \varepsilon_n) \in \{ 0,1 \}^n$. The value of $n$ is the same. Dec 10, 2019 at 11:55