Formal Logic/Sentential Logic/Informal Conventions

= Informal Conventions =

In The Sentential Language, we gave an informal description of a sentential language, namely $$\mathcal{L_S}\,\!$$. We have also given a Formal Syntax for $$\mathcal{L_S}\,\!$$. Our official grammar generates a large number of parentheses. This makes formal definitions and other specifications easier to write, but it makes the language rather cumbersome to use. In addition, all the subscripts and superscripts quickly get to be unnecessarily tedious. The end result is an ugly and difficult to read language.

We will continue to use official grammar for specifying formalities. However, we will informally use a less cumbersome variant for other purposes. The transformation rules below convert official formulae of $$\mathcal{L_S}\,\!$$ into our informal variant.

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


 * The official grammar required sentence letters to have the superscript '0'. Superscripts aren't necessary or even useful until we get to the predicate logic, so we will always omit them in our informal variant.  We will write, for example, $$\mathrm{P_0}\,\!$$ instead of $$\mathrm{P^0_0}\,\!$$.


 * We will omit the subscript if it is '0'. Thus we will write $$\mathrm{P}\,\!$$ instead of $$\mathrm{P^0_0}\,\!$$.  However, we cannot omit all subscripts; we still need to write, for example, $$\mathrm{P_1}\,\!$$.


 * We will omit outermost parentheses. For example, we will write


 * $$\mathrm{P} \rightarrow \mathrm{Q}\,\!$$


 * instead of


 * $$(\mathrm{P^0_0} \rightarrow \mathrm{Q^0_0})\ .\,\!$$


 * We will let a series of the same binary connective associate on the right. For example, we can transform the official


 * $$(\mathrm{P^0_0} \land (\mathrm{Q^0_0} \land \mathrm{R^0_0}))\,\!$$


 * into the informal


 * $$\mathrm{P} \land \mathrm{Q} \land \mathrm{R}\ .\,\!$$


 * However, the best we can do with


 * $$((\mathrm{P^0_0} \land \mathrm{Q^0_0}) \land \mathrm{R^0})\,\!$$


 * is


 * $$(\mathrm{P} \land \mathrm{Q}) \land \mathrm{R}\ .\,\!$$


 * 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


 * $$\mathrm{P} \rightarrow \mathrm{Q} \lor \mathrm{R}\,\!$$


 * instead of


 * $$(\mathrm{P^0_0} \rightarrow (\mathrm{Q^0_0} \lor \mathrm{R^0_0}))\ .\,\!$$


 * However, we cannot remove the internal parentheses from


 * $$((\mathrm{P^0_0} \rightarrow \mathrm{Q^0_0}) \lor \mathrm{R^0_0})\ .\,\!$$


 * Our informal variant of this latter formula is


 * $$(\mathrm{P} \rightarrow \mathrm{Q}) \lor \mathrm{R}\ .\,\!$$


 * Full precedence rankings are given below.

Precedence and scope
Precedence rankings indicate the order that we evaluate the sentential connectives. $$\lor\,\!$$ has a higher precedence than $$\rightarrow\,\!$$. Thus, in calculating the truth value of


 * $$(1) \quad \mathrm{P} \rightarrow \mathrm{Q} \lor \mathrm{R}\ ,\,\!$$

we start by evaluating the truth value of


 * $$(2) \quad \mathrm{Q} \lor \mathrm{R}\,\!$$

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:

Examples
Let's look at some examples. First,


 * $$((\mathrm{P^0_0} \rightarrow \mathrm{Q^0_0}) \leftrightarrow ((\mathrm{P^0_0} \land \mathrm{Q^0_0}) \lor ((\lnot \mathrm{P^0_0} \land \mathrm{Q^0_0}) \lor (\lnot \mathrm{P^0_0} \land \lnot \mathrm{Q^0_0}))))\,\!$$

can be written informally as


 * $$\mathrm{P} \rightarrow \mathrm{Q} \leftrightarrow \mathrm{P} \land \mathrm{Q} \lor \lnot \mathrm{P} \land \mathrm{Q} \lor \lnot \mathrm{P} \land \lnot \mathrm{Q}\ .\,\!$$

Second,


 * $$((\mathrm{P^0_0} \leftrightarrow \mathrm{Q^0_0}) \leftrightarrow ((\mathrm{P^0_0} \land \mathrm{Q^0_0}) \lor (\lnot \mathrm{P^0_0} \land \lnot \mathrm{Q^0_0})))\,\!$$

can be written informally as


 * $$(\mathrm{P} \leftrightarrow \mathrm{Q}) \leftrightarrow \mathrm{P} \land \mathrm{Q} \lor \lnot \mathrm{P} \land \lnot \mathrm{Q}\ .\,\!$$

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


 * $$(\mathrm{P} \rightarrow \mathrm{Q}) \leftrightarrow (\mathrm{P} \land \mathrm{Q}) \lor (\lnot \mathrm{P} \land \mathrm{Q}) \lor (\lnot \mathrm{P} \land \lnot \mathrm{Q})\,\!$$

and


 * $$(\mathrm{P} \leftrightarrow \mathrm{Q}) \leftrightarrow (\mathrm{P} \land \mathrm{Q}) \lor (\lnot \mathrm{P} \land \lnot \mathrm{Q})\ .\,\!$$

Note that the informal formula


 * $$\lnot \mathrm{P} \rightarrow \mathrm{Q}\,\!$$

is restored to its official form as


 * $$(\lnot \mathrm{P^0_0} \rightarrow \mathrm{Q^0_0})\ .\,\!$$

By contrast, the informal formula


 * $$\lnot (\mathrm{P} \rightarrow \mathrm{Q})\,\!$$

is restored to its official form as


 * $$\lnot (\mathrm{P^0_0} \rightarrow \mathrm{Q^0_0})\ .\,\!$$