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First, let me mention that one must be careful when asserting that an implication fails. Taken literally, the assertion "D(x) does not imply P(x)" is logically equivalent to the assertion that D(x) is true and P(x) is false. This meaning of material implication that is used in mathematics is not the same as the natural language interpretation of if-then. For example, if the professor says to a student "It is not true that if you pass the final, then you pass the class", most people would not want the students to deduce logically that he or she will pass the final, but fail the class. But this does follow logically from the mathematical usage of material implication. So your definition of "defect" may not be what you intend.

To be sure, mathematicians are often sloppy about this. One often hears people say that such-and-such condition does not imply another condition. What they mean is that it does not necessarily imply the other condition. For example, suppose I have a function f, and someone says "its not true that if f is continuous, then f is differentiable." This statement is logically equivalent to the assertion that f is indeed continuous and not differentiable. What they meant to say, of course, was that "not every continuous function is differentiable".

In your case, you assert two implication failures: one if the definition of defect and another in the definition of minimal. I believe that when When you clarify exactly what you mean more precisely, you will be led to the conclusion that the only sensible (minimal) defect is simply the assertion D(x), which asserts asserting that "either P(x) holds, or Q(x) fails". This will be the only defectstatement does not imply P(x), except for those values of x for which Q(x) already implies P(x), and it will be minimalalso if D(x) Q(x), since it will be logically equivalent to all then P(x) follows immediately. If D'(x) is any other defectsstatement such that D'(x) Q(x) implies P(x), then D'(x) implies that either Q(x) fails or P(x) holds, and so D'(x) implies D(x).

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First, let me mention that one must be careful when asserting that an implication fails. Taken literally, the assertion "D(x) does not imply P(x)" is logically equivalent to the assertion that D(x) is true and P(x) is false. This meaning of material implication that is used in mathematics is not the same as the natural language interpretation of if-then. For example, if the professor says to a student "It is not true that if you pass the final, then you pass the class", most people would not want the students to deduce logically that he or she will pass the final, but fail the class. But this does follow logically from the mathematical usage of material implication. So your definition of "defect" may not be what you intend. A related issue is involved when there

To be sure, mathematicians are quantifiersoften sloppy about this. When you One often hears people say D that such-and-such condition does not imply P, perhaps you just another condition. What they mean is that there it does not necessarily imply the other condition. For example, suppose I have a function f, and someone says "its not true that if f is at least one x such continuous, then f is differentiable." This statement is logically equivalent to the assertion that D(x) holds but P(x) failsf is indeed continuous and not differentiable. What they meant to say, of course, was that "not every continuous function is differentiable".

In any your case, you assert two implication failures: one if the unique definition of defect and another in the definition of minimal. I believe that when you clarify exactly what you mean more precisely, you will be led to the conclusion that the only sensible defect is simply the assertion D(x) = D(x), which asserts "either P(x) holds, or Q(x) fails". This is a will be the only defectbecause it does not imply P(x), except for those x for which Q(x) already implies P(x), and because if D(x) and Q(x), then P(x) follows. It is it will be minimal, since if D'(x) is any it will be logically equivalent to all other defect, then D'(x) Q(x) implies P(x), and so D'(x) implies either that P(x) holds or Q(x) fails, which is exactly D(x)defects.

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First, let me mention that one must be careful when asserting that an implication fails. Taken literally, the assertion "D(x) does not imply P(x)" is logically equivalent to the assertion that D(x) is true and P(x) is false. This meaning of material implication that is used in mathematics is not the same as the natural language interpretation of if-then. For example, if the professor says to a student "It is not true that if you pass the final, then you pass the class", most people would not want the students to deduce logically that he or she will pass the final, but fail the class. But this does follow logically from the mathematical usage of material implication. So your definition of "defect" may not be what you intend. A related issue is involved when there are quantifiers. When you say D does not imply P, perhaps you just mean that there is at least one x such that D(x) holds but P(x) fails.

In any case, the unique minimal defect is simply the assertion D(x) = "either P(x) holds, or Q(x) fails". This is a defect because it does not imply P(x), except for those x for which Q(x) already implies P(x), and because if D(x) and Q(x), then P(x) follows. It is minimal, since if D'(x) is any other defect, then D'(x) ∧ Q(x) implies P(x), and so D'(x) implies either that P(x) holds or Q(x) fails, which is exactly D(x).