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Additional resources for Aristotle's Modal Syllogisms
Proof. (a) Suppose we have LAbia among the antecedents. Then, since LAbia Aac Abic (Barbara X L X ) , we have Abic LAbia LIac (Darapti X L L ) , so that the expression is asserted. A similar argument applies if the antecedents include LAbic. (b) Identify all variables other than a and c with b. We get no LAba nor LAbc, since no antecedents LAbia or LAbic are present. 51 1 LAab Aca Abc 3 LIac. 5 12. The consequent is LIac, and the antecedents include neither LAaa nor LAcc nor Aac, though they do include Aca.
46 T H E SYSTEM O F A P O D E I C T I C MOODS above the L X L moods in table 8, and so are not implied by any of our axioms. Even if they might have been acceptable to Aristotle, it is unlikely that their L M L and M L L counterparts would have been, as Lukasiewicz himself recognises. Hence we are left with the following valid moods: L X L I X L L , X M M I M X M , L M X I M L X . 3 x 23 = 69 X X X , L X X , X L X , X X M , LLL, LMM, M L M , L L X , L X M , X L M , L L M . . 11 x 24 = 264 Total valid L-X-M moods.
The consequent is LAac, and the antecedents do not include LAcc. If (a) there is LAbic and an Aabi-chain, the expression is asserted. Otherwise (b) it is rejected. Proof. (a) The combination of LAbac and Aabg yields LAac by Barbara L X L . 12. (b) Put bg = b for all LAbic. We get no Aab, since Aabi LAbic => rj LAac, nor Acb, since Acbi LAbic + LAcc. Identify the remainder of the variables with a. We get no LAac, since no LAbic are left, nor Aab, since Abibj LAbjc a LAbgc. 21 LAbb MAab Aac LAca LAbc =- LAac.