(PP) For any property J and any property K, [PJ & [](x)(Jx --> Kx)] --> PK

Since A is a perfection and B is a non perfection:

[Prem] PA & ~PB

Now, assume for reductio that ~<>(Ex)(Ax). This is equivalent to []~(Ex)(Ax) which is equivalent to [](x)~(Ax). Since necessarily A is not exemplified by anything, then trivially [](x)(Ax --> Bx). So by our premise, [PA & [](x)(Ax --> Bx)] But by (PP) [PA & [](x)(Ax --> Bx)] --> PB. It follows that PB. But by our premise, ~PB. This is a contradiction. So we must reject our assumption. So ~~<>(Ex)(Ax). So <>(Ex)(Ax).

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