A implies I

The following argument for the 'existential import' of the universal
proposition just came to mind.

1. Assume 'every car in the street is green' is true.

2. Then 'no car in the street' is green is false, for 'every A is B'
and 'no A is B' are contraries: they cannot be true at the same time.

3. But 'no car in the street is green' is equivalent to 'it is not the
case that some car in the street is green', i.e. is equivalent to the
negation of 'some car in the street is green'.

4. But if it is false, the negation of the negation of 'some car in
the street is green' is true.

5. So 'some car in the street is green' is true.

6. So (1, 5) 'every car in the street is green' implies 'some car in
the street is green'.


Obviously this is not true in predicate logic, if 'every car in the
street is green' is read as, not for some x, x is a car in the street
and x is not green, and if 'no car in the street is green 'not for some
x, x is a car in the street and x is green. For both of these can be
true (if there are no cars in the street). But it does seem odd to say
that every A is B, and no A is B, could both be true.

.

Relevant Pages

  • The suppressed term of negation
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  • Re: The suppressed term of negation
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  • Re: A implies I
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  • Re: A implies I
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