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, 03:46, 17 December 2010
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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}, |
| \\[4pt] | | \\[4pt] |
− | S^{\prime}, S^{\prime\prime}, S^{\prime\prime\prime} ~\operatorname{are}~ q, | + | S^{\prime}, S^{\prime\prime}, S^{\prime\prime\prime} ~\operatorname{are}~ q; |
| \\[4pt] | | \\[4pt] |
| \therefore t ~\operatorname{is}~ q. | | \therefore t ~\operatorname{is}~ q. |
| \end{matrix}</math> | | \end{matrix}</math> |
| </center> | | </center> |
| + | |
| + | <p>Such an argument is double. It combines the two following: |
| + | |
| + | <center> |
| + | <math>\begin{matrix} |
| + | 1. |
| + | \\[4pt] |
| + | S^{\prime}, S^{\prime\prime}, S^{\prime\prime\prime} ~\operatorname{are~taken~as~being}~ P^{\prime}, P^{\prime\prime}, P^{\prime\prime\prime}, |
| + | \\[4pt] |
| + | S^{\prime}, S^{\prime\prime}, S^{\prime\prime\prime} ~\operatorname{are}~ q; |
| + | \\[4pt] |
| + | \therefore ~(\operatorname{By~induction})~ P^{\prime}, P^{\prime\prime}, P^{\prime\prime\prime} ~\operatorname{is}~ q, |
| + | \\[4pt] |
| + | t ~\operatorname{is}~ P^{\prime}, P^{\prime\prime}, P^{\prime\prime\prime}; |
| + | \\[4pt] |
| + | \therefore ~(\operatorname{Deductively})~ t ~\operatorname{is}~ q. |
| + | \end{matrix}</math> |
| + | </center> |
| + | |
| |} | | |} |
| | | |