Last modified on 27 September 2010, at 23:55

# Informal ConventionsEdit

Our predicate language, $\mathcal{L_P}\,\!$, like $\mathcal{L_S}\,\!$, is fairly ugly and difficult to read. The superscripts and subscripts are distracting, and we have multiplied parentheses well beyond necessity. As with sentential logic, will informally use a less cumbersome variant. The official language, however, is still used for definitions formalities and other formalities. Most of the rules for generating informal variants of will be familiar from sentential logic.

## Transformation rulesEdit

We create informal variants of official $\mathcal{L_P}\,\!$ formulae as follows. The examples are cumulative.

• We drop a subscript if it is '0'. Thus we write $a^2\,\!$ instead of $a^2_0\,\!$ and write $\mathrm{F^3}\,\!$ instead of $\mathrm{F^3_0}\,\!$. We cannot drop the subscript from $\mathrm{F^0_1}\,\!$.
• The official grammar requires operation letters and predicate letters to have the superscript indicating the number of places. For example, $a^2_0\,\!$ is a two-place operation letter and $\mathrm{F^3_0}\,\!$ is a three place predicate letter. In most cases, we can drop the superscript and let the context show the number of places. For example, we can write
$a(x, y)\,\!$
Here we observe from the context that the operation letter has two places, thus leaving it understood that $a\,\!$ is an informal variant of $a^2_0\,\!$. Similarly, we observe from the context that the predicate letter in
$\mathrm{F}(x, y, z)\ .\,\!$
has three places. This makes $\mathrm{F}\,\!$, as used in this context, an informal variant of $\mathrm{F^3_0}\,\!$. This convention assumes our $\mathcal{L_P}\,\!$ to be grammatically correct. In general, we will not be trafficking in grammatically incorrect expressions. We will also try to avoid, for example, using $\mathrm{F^1_0}\,\!$ and $\mathrm{F^3_0}\,\!$ in close proximity. Otherwise, their informal variant could cause confusion.
• We will omit outermost parentheses. For example, we will write
$\mathrm{F}(x) \rightarrow \mathrm{G}(y)\,\!$
$(\mathrm{F^1_0}(x_0) \rightarrow \mathrm{G^1_0}(y_0))\ .\,\!$
• We will let a series of the same binary connective associate on the right. For example, we can transform the official
$(\mathrm{F^1_0}(x_0) \land (\mathrm{G^1_0}(x_0) \land \mathrm{H^1_0}(x_0)))\,\!$
into the informal
$\mathrm{F}(x) \land \mathrm{G}(x) \land \mathrm{H}(x)\ .\,\!$
However, the best we can do with
$((\mathrm{F^1_0}(x_0) \land \mathrm{G^1_0}(x_0)) \land \mathrm{H^1_0}(x_0))\,\!$
is
$(\mathrm{F}(x) \land \mathrm{G}(x)) \land \mathrm{H}(x)\ .\,\!$
• We will use precedence rankings to omit internal parentheses when possible. For example, we will regard $\rightarrow\,\!$ as having lower precedence (wider scope) than $\lor\,\!$. This allows us to write
$\mathrm{F}(x) \rightarrow \mathrm{G}(x) \lor \mathrm{H}(x)\,\!$
$(\mathrm{F^1_0}(x_0) \rightarrow (\mathrm{G^1_0}(x_0) \lor \mathrm{H^1_0}(x_0)))\ .\,\!$
However, we cannot remove the internal perentheses from
$((\mathrm{F^1_0}(x_0) \rightarrow \mathrm{G^1_0}(x_0)) \lor \mathrm{H^1_0}(x_0))\ .\,\!$
Our informal variant of this latter formula is
$(\mathrm{F}(x) \rightarrow \mathrm{G}(x)) \lor \mathrm{H}(x)\ .\,\!$

## Precedence and scopeEdit

Precedence rankings indicate the order that we evaluate the sentential connectives and quantifier phrases. $\lor\,\!$ has a higher precedence than $\rightarrow\,\!$. Thus, in evaluating

$(1) \quad \mathrm{F}(x) \rightarrow \mathrm{G}(x) \lor \mathrm{H}(x)\ ,\,\!$

we start by evaluating

$(2) \quad \mathrm{G}(x) \lor \mathrm{H}(x)\,\!$

first. Scope is the length of expression that is governed by the connective. The occurrence of $\rightarrow\,\!$ in (1) has a wider scope than the occurrence of $\lor\,\!$. Thus the occurrence of $\rightarrow\,\!$ in (1) governs the whole sentence while the occurrence of $\lor\,\!$ in (1) governs only the occurrence of (2) in (1).

The full ranking from highest precedence (narrowest scope) to lowest precedence (widest scope) is:

 $\lnot,\ \forall x,\ \exists x\,\!$ highest precedence (narrowest scope) $\land\,\!$ $\lor\,\!$ $\rightarrow\,\!$ $\leftrightarrow\,\!$ lowest precedence (widest scope)

Quantifier phrases have the same precedence as negation signs.

## ExamplesEdit

The following examples are predicate logic variants of the examples at the sentential logic Informal Conventions page. First,

$((\mathrm{F^1_0}(x_0) \rightarrow \mathrm{G^1_0}(x_0)) \leftrightarrow ((\mathrm{F^1_0}(x_0) \land \mathrm{G^1_0}(x_0)) \lor ((\lnot \mathrm{F^1_0}(x_0) \land \mathrm{G^1_0}(x_0)) \lor (\lnot \mathrm{F^1_0}(x_0) \land \mathrm{G^1_0}(x_0))))\,\!$

can be written informally as

$\mathrm{F}(x) \rightarrow \mathrm{G}(x) \leftrightarrow \mathrm{F}(x) \land \mathrm{G}(x) \lor \lnot \mathrm{F}(x) \land \mathrm{G}(x) \lor \lnot \mathrm{F}(x) \land \mathrm{G}(x)\ .\,\!$

Second,

$((\mathrm{F^1_0}(x_0) \leftrightarrow \mathrm{G^1_0}(x_0)) \leftrightarrow ((\mathrm{F^1_0}(x_0) \land \mathrm{G^1_0}(x_0)) \lor (\lnot \mathrm{F^1_0}(x_0) \land \lnot \mathrm{G^1_0}(x_0))))\,\!$

can be written informally as

($\mathrm{F}(x) \leftrightarrow \mathrm{G}(x)) \leftrightarrow \mathrm{F}(x) \land \mathrm{G}(x) \lor \lnot \mathrm{F}(x) \land \lnot \mathrm{G}(x)\ .\,\!$

Some unnecessary parentheses may prove helpful. In the two examples above, the informal variants may be easier to read as

$(\mathrm{F}(x) \rightarrow \mathrm{G}(x)) \leftrightarrow (\mathrm{F}(x) \land \mathrm{G}(x)) \lor (\lnot \mathrm{F}(x) \land \mathrm{G}(x)) \lor (\lnot \mathrm{F}(x) \land \mathrm{G}(x))\,\!$

and

$(\mathrm{F}(x) \leftrightarrow \mathrm{G}(x)) \leftrightarrow (\mathrm{F}(x) \land \mathrm{G}(x)) \lor (\lnot \mathrm{F}(x) \land \lnot \mathrm{G}(x))\ .\,\!$

Note that the informal formulae

$\lnot \mathrm{F}(x) \rightarrow \mathrm{G}(x)\,\!$
$\forall x \mathrm{F}(x) \rightarrow \mathrm{G}(x)\,\!$
$\exists x \mathrm{F}(x) \land \mathrm{G}(x)\,\!$

are restored to its official form as

$(\lnot \mathrm{F^1_0}(x_0) \rightarrow \mathrm{F^1_0}(x_0))\ .\,\!$
$(\forall x_0 \mathrm{F^1_0}(x_0) \rightarrow \mathrm{F^1_0}(x_0))\ .\,\!$
$(\exists x_0 \mathrm{F^1_0}(x_0) \land \mathrm{F^1_0}(x_0))\ .\,\!$

By contrast, the informal formula

$\lnot (\mathrm{F}(x) \rightarrow \mathrm{G}(x))\,\!$
$\forall x (\mathrm{F}(x) \rightarrow \mathrm{G}(x))\,\!$
$\exists x (\mathrm{F}(x) \land \mathrm{G}(x))\,\!$

is restored to its official form as

$\lnot (\mathrm{F^1_0}(x_0) \rightarrow \mathrm{G^1_0}(x_0))\ .\,\!$
$\forall x_0 (\mathrm{F^1_0}(x_0) \rightarrow \mathrm{G^1_0}(x_0))\ .\,\!$
$\exists x_0 (\mathrm{F^1_0}(x_0) \land \mathrm{G^1_0}(x_0))\ .\,\!$