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Proof 1 proceeded by the ''straightforward approach'', starting with ''e''₂ as ''s''₁ and ending with ''e''₃ as ''s''_''n''. That is, it commenced from the sign "(p (q))(p (r))" and ended up at the sign "(p (q r))" 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 ''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.