In a square of opposition, if 'No animals with horns are carnivores' is false, which of the following is the correct code?
codes:
A. 'Some animals with horns are carnivores' is true.
B. 'Some animals with horns are not carnivores' is true.
C. 'All animals with horns are carnivores' is false
D. 'All the animals with horns are carnivores' is undermined.
A and D Only
In the traditional square of opposition, the four standard-form propositions about a subject-predicate pair are: A (universal affirmative, 'All S are P'), E (universal negative, 'No S are P'), I (particular affirmative, 'Some S are P'), and O (particular negative, 'Some S are not P').
Here S = 'animals with horns' and P = 'carnivores', and we are told the E proposition 'No animals with horns are carnivores' is false.
E and I are contradictories: they cannot both be true and cannot both be false - exactly one of them must be true. Since E is false, its contradictory I ('Some animals with horns are carnivores') must be true. This confirms code A.
A and E are contraries: they cannot both be true, but they can both be false. Knowing only that E is false does not by itself fix the truth value of A ('All animals with horns are carnivores') - it could be true or it could be false. So the truth value of the A proposition remains logically undetermined, which confirms code D.
I and O are subcontraries: they cannot both be false, but they can both be true. Since I has already been established as true, this relation places no constraint that would make O ('Some animals with horns are not carnivores') definitely true - O's truth value is likewise left undetermined, so code B (claiming O is definitely true) is incorrect.
Code C claims the A proposition is definitely false, but as shown above the contrary relation only tells us A and E cannot both be true; it does not force A to be false whenever E is false, so code C is also incorrect.
Therefore, only codes A and D validly follow from the falsity of the E proposition, making A and D only the correct answer.
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