Changes

==Analysis of contingent propositions==

==Analysis of contingent propositions==

For all of the reasons mentioned above, and for the sake of a more compact illustration of the in and outs of a typical propositional equation reasoning system, let's now take up a much simpler example of a contingent proposition:

For all of the reasons mentioned above, and for the sake of a more compact illustration of the ins and outs of a typical propositional equation reasoning system, let's now take up a much simpler example of a contingent proposition:

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I know that it must seem tedious, but I probably ought to go ahead and carry out the second half of this analogically model-theoretic strategy, just so that we will have the security of this concrete and shared experience on which to fall back at every later point in what may quickly become a rather abstruse discussion.  Here then is the rest of the necessary chain of equations:

We probably ought to go ahead and carry out the second half of this model-theoretic strategy, just so we'll have the security of this concrete experience to call on in future discussions.  The remainder of the needed chain of equations is as follows:

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This is not only a logically equivalent DNF but exactly the same DNF expression that we obtained before, so we have established the given equation <math>\texttt{(} p \texttt{(} q \texttt{))(} p \texttt{(} r \texttt{))} = \texttt{(} p \texttt{(} q r \texttt{))}.</math>

This is not only a logically equivalent DNF but exactly the same DNF expression that we obtained before, so we have established the given equation <math>\texttt{(} p \texttt{(} q \texttt{))(} p \texttt{(} r \texttt{))} = \texttt{(} p \texttt{(} q r \texttt{))}.</math> Incidentally, one may wish to note that this DNF expression quickly folds into the following form:

 

Incidentally, one may wish to note that this DNF expression quickly folds into the following form:

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This can be read to say <math>{}^{\backprime\backprime} \operatorname{either}~ p q r, ~\operatorname{or}~ \operatorname{not}~ p {}^{\prime\prime},</math> which gives us yet another equivalent to the expression <math>\texttt{(} p \texttt{(} q \texttt{))(} p \texttt{(} r \texttt{))}</math> and the expression <math>\texttt{(} p \texttt{(} q r \texttt{))}.</math>  Still another way of writing the same thing would be as follows:

This can be read to say <math>{}^{\backprime\backprime} \operatorname{either}~ p q r ~\operatorname{or}~ \operatorname{not}~ p {}^{\prime\prime},</math> which gives us yet another equivalent for the expression <math>\texttt{(} p \texttt{(} q \texttt{))(} p \texttt{(} r \texttt{))}</math> and the expression <math>\texttt{(} p \texttt{(} q r \texttt{))}.</math>  Still another way of writing the same thing would be as follows:

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This one is easy enough to derive from rules that are already known, but call it the ''Indistinctness Rule'' just on behalf of ready reference and employment.

This one is easy enough to derive from rules that are already known, but just for the sake of ready reference it is useful to canonize it as the ''Indistinctness Rule''. Finally, let me introduce a rule-of-thumb that is a bit more suited to routine computation, and that serves to replace the indistinctness rule in many cases where we actually have to call on it.  This is actually just a special case of the evaluation rule listed above:

 

Finally, let me introduce a rule-of-thumb that is a bit more suited to routine computation, and that will serve to replace the indistinctness rule in many of the cases where we actually have to call on it.  This is actually just a special case of the evaluation rule listed above:

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To continue with the beating of this still kicking horse that is known as the propositional equation "(p (q))(p (r)) = (p (q r))", let's now take up the third way that I mentioned for examining propositional equations, but I believe that you will be relieved to know that it is literally a third way only at the very outset, almost immediately breaking up according to whether one proceeds by way of the more "routine" model-theoretic path or else by way of the more "strategic" proof-theoretic path.  I think that I'll take the low road today.

To continue with the beating of this still-kicking horse in the form of the equation <math>\texttt{(} p \texttt{(} q \texttt{))(} p \texttt{(} r \texttt{))} = \texttt{(} p \texttt{(} q r \texttt{))},</math> let's now take up the third way that I mentioned for examining propositional equations, even if it is literally a third way only at the very outset, almost immediately breaking up according to whether one proceeds by way of the more routine model-theoretic path or else by way of the more strategic proof-theoretic path.

Let's convert the equation between propositions:

Let's convert the equation between propositions:

: (p (q))(p (r)) = (p (q r)),

\texttt{(} p \texttt{(} q \texttt{))(} p \texttt{(} r \texttt{))}

& = &

\texttt{(} p \texttt{(} q r \texttt{))}

into the corresponding equational proposition:

into the corresponding equational proposition:

: (( (p (q))(p (r)) , (p (q r)) )).

\texttt{((}

\texttt{(} p \texttt{(} q \texttt{))(} p \texttt{(} r \text{))}

& \texttt{,} &

\texttt{(} p \texttt{(} q r \texttt{))}

\texttt{))}

If you're like me, you'd rather see it in pictures:

If you're like me, you'd rather see it in pictures: