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Revision as of 16:28, 21 August 2009
, 16:28, 21 August 2009→Proof as semiosis: del redun
: This way of proceeding went from <math>e_5 = {}^{\backprime\backprime} \texttt{((~(} p \texttt{~(} q \texttt{))~(} p \texttt{~(} r \texttt{))~,~(} p \texttt{~(} q~r \texttt{))~))} {}^{\prime\prime}</math> to the blank expression that <math>\operatorname{Ex}</math> recognizes as the value <math>\operatorname{true}.</math>
: This way of proceeding went from <math>e_5 = {}^{\backprime\backprime} \texttt{((~(} p \texttt{~(} q \texttt{))~(} p \texttt{~(} r \texttt{))~,~(} p \texttt{~(} q~r \texttt{))~))} {}^{\prime\prime}</math> to the blank expression that <math>\operatorname{Ex}</math> recognizes as the value <math>\operatorname{true}.</math>
==Computation and inference as semiosis==
==Computation and inference as semiosis==
