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t ~\operatorname{is}~ P^{\prime}, P^{\prime\prime}, P^{\prime\prime\prime},
t ~\operatorname{is}~ P^{\prime}, P^{\prime\prime}, P^{\prime\prime\prime},
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\\[4pt]
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S^{\prime}, S^{\prime\prime}, S^{\prime\prime\prime} ~\operatorname{are}~ q,
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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.
\end{matrix}</math>
\end{matrix}</math>
</center>
</center>
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<p>Such an argument is double. It combines the two following:
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<center>
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<math>\begin{matrix}
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1.
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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},
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\\[4pt]
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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,
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\\[4pt]
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t ~\operatorname{is}~ P^{\prime}, P^{\prime\prime}, P^{\prime\prime\prime};
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\\[4pt]
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\therefore ~(\operatorname{Deductively})~ t ~\operatorname{is}~ q.
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\end{matrix}</math>
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</center>
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