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Revision as of 03:00, 21 August 2009
, 03:00, 21 August 2009→Proof as semiosis: TeX markup
Under <math>\operatorname{Ex}</math> we have the following interpretations:
Under <math>\operatorname{Ex}</math> we have the following interpretations:
{| align="center" cellpadding="8" width="90%"
| <math>e_0 = {}^{\backprime\backprime} \texttt{(~)} {}^{\prime\prime}</math> expresses the logical constant <math>\operatorname{false}.</math>
|-
| <math>e_1 = {}^{\backprime\backprime} \texttt{~} {}^{\prime\prime}</math> expresses the logical constant <math>\operatorname{true}.</math>
|-
| <math>e_2 = {}^{\backprime\backprime} \texttt{(} p \texttt{~(} q \texttt{))~(} p \texttt{~(} r \texttt{))} {}^{\prime\prime}</math> says <math>{}^{\backprime\backprime} \operatorname{not}~ p ~\operatorname{without}~ q,</math> <math>\operatorname{and~not}~ p ~\operatorname{without}~ r {}^{\prime\prime}.</math>
|-
| <math>e_3 = {}^{\backprime\backprime} \texttt{(} p \texttt{~(} q~r \texttt{))} {}^{\prime\prime}</math> says <math>{}^{\backprime\backprime} \operatorname{not}~ p ~\operatorname{without}~ q ~\operatorname{and}~ r {}^{\prime\prime}.</math>
|-
| <math>e_4 = {}^{\backprime\backprime} \texttt{(} p~q~r \texttt{~,~(} p \texttt{))} {}^{\prime\prime}</math> says <math>{}^{\backprime\backprime} p ~\operatorname{and}~ q ~\operatorname{and}~ r,</math> <math>~\operatorname{or~else~not}~ p{}^{\prime\prime}.</math>
|-
| <math>e_5 = {}^{\backprime\backprime} \texttt{((~(} p \texttt{~(} q \texttt{))~(} p \texttt{~(} r \texttt{))~,~(} p \texttt{~(} q~r \texttt{))~))} {}^{\prime\prime}</math> says that <math>e_2\!</math> and <math>e_3\!</math> say the same thing.
|}
We took up the Equation <math>E_1\!</math> that reads as follows:
We took up the Equation ''E''<sub>1</sub> that reads as follows:
: (p (q))(p (r)) = (p (q r)).
: (p (q))(p (r)) = (p (q r)).
