Tuesday, November 20, 2007

6.1203 In order to perceive a tautology as such, one can, in cases in which no sign of generality occurs in the tautology, avail oneself of the following method: I write “TpF”, “TqF”, “TrF”, etc. in stead of “p”, “q”, “r”, etc. I express the truth-combinations with brackets, e.g.:

diagram of p/q=F/F F/T T/F T/T

The coordination of the truth or falsity of the whole proposition and the truth-combinations of the truth-arguments with lines in the following way:

T (T/F)->F" shapes="_x0000_i1026" height="96" width="128">

This sign, e.g., would therefore present the proposition p → q. Now I will investigate on the strength of that whether, e.g., the proposition ~(p. ~p) (the Law of Contradiction) is a tautology. The form “~ξ” gets written in our notation as;

T, (T)->F" shapes="_x0000_i1029" height="64" width="64">T, (T)->F" shapes="_x0000_i1029" height="64" width="64">T, (T)->F" shapes="_x0000_i1029" height="64" width="64">

the form “ξ . η” thus:

F (T/T)->T" shapes="_x0000_i1032">

So the proposition ~(p. ~q) goes thus:

T, T/F->F" shapes="_x0000_i1033">T, T/F->F" shapes="_x0000_i1033">T, T/F->F" shapes="_x0000_i1033">

If we put here “p” instead of “q” and investigate the combination of the outermost T and F with the innermost, then we get the result that the truth of the whole proposition is coordinated with all the truth-combinations of its arguments, its falsity with none of the truth-combinations.

A somewhat complicated way of demonstrating an obvious truth, but I suppose it’s the demonstration that matters here. Or maybe the point is that this is all that logic can be/do.

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