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Revision as of 05:01, 26 December 2010
, 05:01, 26 December 2010→Version 1. “On the Natural Classification of Arguments” (1867): modify notation
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<math>\begin{matrix}
T ~\operatorname{is}~ P^{\prime}, P^{\prime\prime}, P^{\prime\prime\prime},
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S^{\prime}, S^{\prime\prime}, ~\operatorname{and}~ S^{\prime\prime\prime} ~\operatorname{are}~ q;
S^{\prime}, S^{\prime\prime}, S^{\prime\prime\prime} ~\operatorname{are}~ Q;
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\therefore t ~\operatorname{is}~ q.
\therefore T ~\operatorname{is}~ Q.
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S^{\prime}, S^{\prime\prime}, S^{\prime\prime\prime} ~\operatorname{are~taken~as~being}~ P^{\prime}, P^{\prime\prime}, P^{\prime\prime\prime},
S^{\prime}, S^{\prime\prime}, S^{\prime\prime\prime} ~\operatorname{are~taken~as~being}~ P^{\prime}, P^{\prime\prime}, P^{\prime\prime\prime},
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S^{\prime}, S^{\prime\prime}, S^{\prime\prime\prime} ~\operatorname{are}~ q;
S^{\prime}, S^{\prime\prime}, S^{\prime\prime\prime} ~\operatorname{are}~ Q;
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\therefore ~(\operatorname{By~induction})~ P^{\prime}, P^{\prime\prime}, P^{\prime\prime\prime} ~\operatorname{is}~ q,
\therefore ~(\operatorname{By~induction})~ P^{\prime}, P^{\prime\prime}, P^{\prime\prime\prime} ~\operatorname{is}~ Q,
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T ~\operatorname{is}~ P^{\prime}, P^{\prime\prime}, P^{\prime\prime\prime};
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\therefore ~(\operatorname{Deductively})~ t ~\operatorname{is}~ q.
\therefore ~(\operatorname{Deductively})~ T ~\operatorname{is}~ Q.
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S^{\prime}, S^{\prime\prime}, S^{\prime\prime\prime} ~\operatorname{are,~for~instance,}~ P^{\prime}, P^{\prime\prime}, P^{\prime\prime\prime},
S^{\prime}, S^{\prime\prime}, S^{\prime\prime\prime} ~\operatorname{are,~for~instance,}~ P^{\prime}, P^{\prime\prime}, P^{\prime\prime\prime},
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T ~\operatorname{is}~ P^{\prime}, P^{\prime\prime}, P^{\prime\prime\prime};
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\therefore ~(\operatorname{By~hypothesis})~ t ~\operatorname{has~the~common~characters~of}~ S^{\prime}, S^{\prime\prime}, S^{\prime\prime\prime},
\therefore ~(\operatorname{By~hypothesis})~ T ~\operatorname{has~the~common~characters~of}~ S^{\prime}, S^{\prime\prime}, S^{\prime\prime\prime},
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S^{\prime}, S^{\prime\prime}, S^{\prime\prime\prime} ~\operatorname{are}~ q;
S^{\prime}, S^{\prime\prime}, S^{\prime\prime\prime} ~\operatorname{are}~ Q;
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\therefore ~(\operatorname{Deductively})~ t ~\operatorname{is}~ q.
\therefore ~(\operatorname{Deductively})~ T ~\operatorname{is}~ Q.
\end{matrix}</math>
\end{matrix}</math>
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