Changes

{| align="center" cellpadding="8" width="90%"

{| align="center" cellpadding="8" width="90%"

| <math>\texttt{(} p \texttt{~(} q \texttt{))(} p \texttt{~(} r \texttt{))} = \texttt{(} p \texttt{~(} q~r \texttt{))}.\!</math>

| <math>\texttt{(} p \texttt{ (} q \texttt{))(} p \texttt{ (} r \texttt{))} = \texttt{(} p \texttt{ (} q~r \texttt{))}.\!</math>

|}

|}

'''Proof&nbsp;1''' proceeded by the ''straightforward approach'', starting with <math>e_2\!</math> as <math>s_1\!</math> and ending with <math>e_3\!</math> as <math>s_n\!.</math>

'''Proof&nbsp;1''' proceeded by the ''straightforward approach'', starting with <math>e_2\!</math> as <math>s_1\!</math> and ending with <math>e_3\!</math> as <math>s_n\!.</math>

: That is, Proof&nbsp;1 commenced from the sign <math>{}^{\backprime\backprime} \texttt{(} p \texttt{~(} q \texttt{))~(} p \texttt{~(} r \texttt{))} {}^{\prime\prime}\!</math> and ended up at the sign <math>{}^{\backprime\backprime} \texttt{(} p \texttt{~(} q~r \texttt{))} {}^{\prime\prime}\!</math> by legal moves.

: That is, Proof&nbsp;1 commenced from the sign <math>{}^{\backprime\backprime} \texttt{(} p \texttt{ (} q \texttt{)) (} p \texttt{ (} r \texttt{))} {}^{\prime\prime}\!</math> and ended up at the sign <math>{}^{\backprime\backprime} \texttt{(} p \texttt{ (} q~r \texttt{))} {}^{\prime\prime}\!</math> by legal moves.

'''Proof&nbsp;2''' lit on by ''burning the candle at both ends'', changing <math>e_2\!</math> into a normal form that reduced to <math>e_4,\!</math> and changing <math>e_3\!</math> into a normal form that also reduced to <math>e_4,\!</math> in this way tethering <math>e_2\!</math> and <math>e_3\!</math> to a common stake.

'''Proof&nbsp;2''' lit on by ''burning the candle at both ends'', changing <math>e_2\!</math> into a normal form that reduced to <math>e_4,\!</math> and changing <math>e_3\!</math> into a normal form that also reduced to <math>e_4,\!</math> in this way tethering <math>e_2\!</math> and <math>e_3\!</math> to a common stake.

: Filling in the details, one route went from <math>{}^{\backprime\backprime} \texttt{(} p \texttt{~(} q \texttt{))~(} p \texttt{~(} r \texttt{))} {}^{\prime\prime}\!</math> to <math>{}^{\backprime\backprime} \texttt{(} p~q~r \texttt{~,~(} p \texttt{))} {}^{\prime\prime},\!</math> and another went from <math>{}^{\backprime\backprime} \texttt{(} p \texttt{~(} q~r \texttt{))} {}^{\prime\prime}\!</math> to <math>{}^{\backprime\backprime} \texttt{(} p~q~r \texttt{~,~(} p \texttt{))} {}^{\prime\prime},\!</math> thus equating the two points of departure.

: Filling in the details, one route went from <math>{}^{\backprime\backprime} \texttt{(} p \texttt{ (} q \texttt{)) (} p \texttt{ (} r \texttt{))} {}^{\prime\prime}\!</math> to <math>{}^{\backprime\backprime} \texttt{(} p~q~r \texttt{ , (} p \texttt{))} {}^{\prime\prime},\!</math> and another went from <math>{}^{\backprime\backprime} \texttt{(} p \texttt{ (} q~r \texttt{))} {}^{\prime\prime}\!</math> to <math>{}^{\backprime\backprime} \texttt{(} p~q~r \texttt{ , (} p \texttt{))} {}^{\prime\prime},\!</math> thus equating the two points of departure.

'''Proof&nbsp;3''' took the path of reflection, expressing the meta-equation between <math>e_2\!</math> and <math>e_3\!</math> in the form of the naturalized equation <math>e_5,\!</math> then taking <math>e_5\!</math> as <math>s_1\!</math> and exchanging it by dint of value preserving steps for <math>e_1\!</math> as <math>s_n.\!</math>

'''Proof&nbsp;3''' took the path of reflection, expressing the meta-equation between <math>e_2\!</math> and <math>e_3\!</math> in the form of the naturalized equation <math>e_5,\!</math> then taking <math>e_5\!</math> as <math>s_1\!</math> and exchanging it by dint of value preserving steps for <math>e_1\!</math> as <math>s_n.\!</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>\mathrm{Ex}\!</math> recognizes as the value <math>{\mathrm{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>\mathrm{Ex}\!</math> recognizes as the value <math>{\mathrm{true}}.\!</math>

==Computation and inference as semiosis==

==Computation and inference as semiosis==