Changes

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, it 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;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, it 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 ''e''₂ into a normal form that reduced to ''e''₄, and changing ''e''₃ into a normal form that also reduced to ''e''₄, in this way tethering ''e''₂ and ''e''₃ to a common stake.  In more detail, one route went from "(p (q))(p (r))" to "(p q r, (p))", and another went from "(p (q r))" to "(p q r, (p))", thus equating the two points of departure.

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.  In more detail, 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 met-equation between ''e''₂ and ''e''₃ in the naturalized equation ''e''₅, then taking ''e''₅ as ''s''₁ and exchanging it by dint of value preserving steps for ''e''₁ as ''s''_''n''.  Thus we went from "(( (p (q))(p (r)) , (p (q r)) ))" to the blank expression that <math>\operatorname{Ex}</math> recognizes as the value ''true''.

Proof&nbsp;3 took the path of reflection, expressing the met-equation between ''e''₂ and ''e''₃ in the naturalized equation ''e''₅, then taking ''e''₅ as ''s''₁ and exchanging it by dint of value preserving steps for ''e''₁ as ''s''_''n''.  Thus we went from "(( (p (q))(p (r)) , (p (q r)) ))" to the blank expression that <math>\operatorname{Ex}</math> recognizes as the value ''true''.