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This inference is only materially valid. It’s form is something like:

If you add the premise that children of U.S. citizens are also U.S. citizens, the inference becomes deductively valid.

This inference is logically valid. As it has the same form as the Socrates inference.

This inference is only materially valid. It’s form is something like:

But if c is little Jimmy, P is hungry, and Q thirsty, the inference becomes: Little Jimmy is not hungry. So, Little Jimmy is thirsty, which is clearly invalid.

We need to add the premise that a sentence is false if and only if it’s not true as a further premise to make this inference valid. This premise is false, for example, in many-valued logic.

This inference is only materially valid, as it has the form c is P. Therefore c is Q. Most c’s, Ps and Qs are counterexamples, but let’s take c to be little Jimmy, P having a blue hat, and Q being an adult. The inference becomes: Little Jimmy has a blue hat. Therefore, he’s an adult.

We need to add the premise that only thinking things exist.

This inference is (interestingly) deductively valid. It is an instance of “modus tolens”: All A’s are B’s and you’re not a B. Therefore, you’re not an A.

The mistake is to forget that there are other unis in the UK that Mr. Sir could have gone to, such as St Andrews for example. In the possibility that he did, the premise is true (he neither went to Oxford nor Cambridge, but the conclusion isn’t—St Andrews is in Scotland, thus the U.K.).

The mistake is to miss that there could be New Yorkers, like little Jimmy, who aren’t Yankees fans. So, if we assume that he is from NY but not a Yankees fan, it could still be the case that only New Yorkers are Yankees fans.

The inference only ever has a chance of being inductively valid, its not even that. The mistake is that the sampling size, though large, is biased. It could be that East Coast burger joints are biased against vegetarians, while the East Coast ones are more open minded. In such a possible situation, the truth of the premise doesn’t make the conclusion likely at all.

The mistake is to forget that systems could be hybrid, that is both statistics-based and logic-based. So, in a possible situation, where there are only the two paradigms (statistics and logic-based), but IA is a hybrid AI system, the premise is true but the conclusion isn’t.

This is another inductive fallacy, known as the Gambler’s fallacy . The mistake is to forget that each spin of the wheel is an independent event, the outcome of any former spin: the chance remains untouched, it still is 1/2 even in situations, where the wheel has spun black many, many times.

{2, 4, 6, 8 } or simply { n : n is an even integer between 1 and 10 }

{ }. This is the empty set again, which has no members, whatsoever.

This claim is false. In fact, we have that since the only members of the first set, Mr. Sir and little Jimmy, are both members of the second set as well.

This claim is true, since both members of the first set—my beer and the number one—are members of the second set. In fact, both sets are the same!

This claim is very importantly false. The empty set, { }, cannot fail to be a subset of or any set for that matter. Because for that to be the case, there would have to be a member of { }, which fails to be a member of . But which member of { } could that be, as there are none.

Since both the soda and the beer are sparkling beverages, they are members of the set { x : x is a sparkling beverage }

There is a total of 8 subsets. Note that the empty set, { }, is among them, according to number 8.:

Formally, we can state the claim as follows:

Applying the definition of deductive consequence in terms of subset and intersection, gives us:

To show the theorem, we can proceed as follows:

• We assume that , since we want to show that if this is true, then is true.

• To show that , we think about what a member of can look like. As we’ll see, just by looking at the definition of , we can see that all elements of this set need to be in T.

• By definition, . That means that any member of is both a member of S and of S’.

• But we’ve assumed that , which means that every member of S is a member of T. And any member of is a member of S. So, every member of is a member of T.

• But that just means that . So, if we assume that , , get , which is the content of our theorem.

The monotonicity of deductive inference is a corollary of our theorem. Just interpret S,S’, and T as follows:

From this observation, our main claim directly follows.

Here’s the calculation:

In this case, something interesting happens:

Note that the intersection of [2] and [EVEN] ∩ [>3] is empty: the result being two is incompatible with being an even result bigger than 3. This is why the result of the calculation is 0.

The definition of inductive validity requires that for each assignment of probabilities, the probability of the conclusion to go up conditional on the premises. But here’s a probability assignment, where the probability of the conclusion goes down: Pr([2] | [EVEN] ∩ [>3]) = 0, which is less than Pr([2]) = 1/6. So the inference is inductively invalid.

What’s going on here is that we’ve added additional information that contradicts our conclusion. That this can happen is characteristic of inductive inference: we can have evidence, which makes our conclusion more likely but is not conclusive, in the sense it still allows for the conclusion to be false.

This generalizes to all inductively valid inferences, which are not also deductively valid. We can always take information that contradicts our conclusion—if you’re hard pressed to find something contradictory, just take the negation of the conclusion—and add it to the premises, and we’ll have lowered the conditional probability of the conclusion given these premises to 0. Whatever the probability of the conclusion was before, if it was at all possible (meaning ≥ 0), it’s now smaller.