Valid inference

Solutions

1. This inference is only materially valid. Its form is something like:

   c's P's are Q, so c is Q.

   But suppose that c is little Jimmy, his P’s are his favorite socks, and Q is being blue. Then the inference becomes clearly invalid: Little Jimmy's favorite socks are blue, so Jimmy is blue.

   If you add the premise that children of club members are also members, the inference becomes logically deductively valid.

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

3. This inference is only materially valid. Its form is something like:

   c is not P. Thus c is Q

   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 need not hold in a many-valued semantics. Here we assume the sentence has exactly one of the two truth-values, true and false.

4. 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.

   On this reading, we need to add the premise that everything that thinks exists. We’re treating existence as a predicate here. In standard FOL, names already denote objects, so expressing the philosophical argument requires care about what counts as an object in the model.

5. This inference is logically deductively valid. It is an instance of "modus tollens": All A's are B's and you're not a B. Therefore, you're not an A.

Solution

1. 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.

2. 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.

3. This isn’t deductively valid: the East Coast observations are compatible with the West Coast restaurant having vegetarian options. Its inductive strength depends on further assumptions. Were the sampled restaurants representative, and are the two populations similar in the relevant way? If menus differ systematically by region, the transfer may fail. The premise alone doesn’t tell us how much support the conclusion receives.

4. 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 ∀I is a hybrid AI system, the premise is true but the conclusion isn’t.

5. This is another inductive fallacy, known as the Gambler’s fallacy (opens in a new tab) . The mistake is to overlook the stated independence assumption. Earlier spins do not change the next spin’s chance of red: it remains 1/2. Our wheel has only red and black outcomes; an ordinary roulette wheel also has zero pockets.

Suggested answers

What would count as checking?

1. Deadline: compare the answer with the relevant passage, including which application, year, and conditions it concerns. For a straightforward report, this may settle the question without inspecting an explanation.
2. Eligibility: verify the applicant’s details and the applicable rules, then check whether the conclusion follows. Correct facts can still be combined incorrectly, for example by confusing necessary and sufficient conditions or overlooking an exception.
3. Timetable: check the proposed timetable against every stated constraint. If it satisfies them all, it is a solution, regardless of how the model found it. A further claim that it is the best timetable needs a separate justification, as do assumptions that the constraints adequately describe the real scheduling problem.
4. Teaching method: verify the report about the earlier class, then investigate what connects those results to your class. Differences between students, teaching conditions, or how success was measured may affect the inference. The future outcome cannot yet be checked against a record.

The appropriate check depends on the claim and what you need it for. Reading an explanation can help expose an invalid step or a hidden assumption, but checking a proposed solution directly can sometimes establish what you need.

Every fact checks out

1. No. The inference has the form “All As are Bs; Sam is B; therefore Sam is A.” This reverses the direction of the first premise. An enrolled student without a scholarship is a counterexample. In conditional form, this is affirming the consequent.
2. You have independently established that the conclusion is true, along with the premises. That does not establish the validity of the presented inference.
3. Add “Every enrolled student receives a scholarship.” Together with Sam’s enrollment, this entails the conclusion. Whether this new premise is true is a further question.
4. The same pattern could wrongly classify enrolled students who do not have scholarships. Success in Sam’s case does not justify using it as a general decision rule. A particular answer can be correct even when the proposed reasoning would fail in other cases.

Can we reuse the reasoning?

For Ben, one compatible situation gives him a valid pass and no suspension, so rule 2 permits entry. Another makes him suspended, so rule 3 prohibits it. For Eli, a valid pass would establish permission. With no valid pass, the rules allow a situation in which he is not permitted to enter—but do not require that outcome. Thus neither permission nor prohibition follows from Eli’s listed facts alone.

Replacing Ada’s non-suspension with suspension establishes prohibition by rule 3. Withholding her pass information instead leaves the case undecided: failing to establish permission is different from establishing prohibition.

A model can reach the right classification with a mistaken justification. For example, saying Ada may enter simply because she is trained would overlook two conditions in rule 2. Correct classifications and valid justifications across these cases are evidence of success on these tests, not a guarantee for every possible rulebook or case.

What assumption is doing the work?

The observation concerns a group of students who used flashcards. It does not by itself establish that flashcards improved their chances, or that the same recommendation applies to you.

In one situation, the exams and students are comparable to yours, and further evidence shows that flashcard practice helped students learn the tested material. In another, already well-prepared students chose flashcards, most would have passed anyway, and your exam tests skills that their cards did not address. The reported observation can be true in both situations, while the case for the recommendation differs.

Your reflection should identify a particular use and a concrete check. For example, if you use AI to plan exam preparation, verify what the exam covers and assess whether the proposed activities actually practice those skills. Another chatbot’s agreement does not by itself settle either question: it may repeat the same assumption or error. Explain what independent evidence or reasoning would give you grounds to rely on the recommendation.

Solutions

1. { }
2. { , }

3. {2, 4, 6, 8}.

4. The empty set, ∅: none of the pictured members occurs in both sets.

5. { }

6. False. Both Mr. Sir and Jimmy belong to the set on the right, so the first set is a subset of it.

7. True. The two sets have exactly the same members; their order does not matter.

8. False. The empty set is a subset of every set: it has no member that could fail to belong to the other set.

9. True under the stated assumption: both objects are sparkling beverages.

10. There are eight subsets, including the empty set and the whole set:

    {} { } { } { } { , } { , } { , } { , , }

Solution

A countermodel must make all premises true and the conclusion false. A false premise therefore disqualifies a world, even if the conclusion is false too.

Solutions

1. If P₁, …, Pₙ ⊨ C, then P₁, …, Pₙ, Q ⊨ C.
2. If [P₁] ∩ … ∩ [Pₙ] ⊆ [C], then [P₁] ∩ … ∩ [Pₙ] ∩ [Q] ⊆ [C].
3. Suppose S ⊆ T. Let x be any member of S ∩ S′. Then x ∈ S and x ∈ S′. Since every member of S belongs to T, we have x ∈ T. Thus every member of S ∩ S′ is in T, which means S ∩ S′ ⊆ T. This also covers an empty intersection: there is then no member that could fail to belong to T.
4. Put S = [P₁] ∩ … ∩ [Pₙ], S′ = [Q], and T = [C]. The theorem gives exactly the claim in 2. Adding a premise restricts the models where all premises are true; it cannot introduce a countermodel.
5. From “All humans are mortal” and “Socrates is human,” infer “Socrates is mortal.” Add “It is raining.” Every model of all three premises still satisfies the first two, and therefore still makes the conclusion true.

Solution

1. Pr([TWO] | [EVEN]) = (1/6) / (3/6) = 1/3, compared with Pr([TWO]) = 1/6. The premise raises the conclusion’s probability.
2. Since [TWO] ⊆ [EVEN], we have Pr([TWO] | [EVEN]) = Pr([TWO]) / Pr([EVEN]) ≥ Pr([TWO]) whenever Pr([EVEN]) > 0. Dividing a nonnegative number by a positive number at most 1 cannot make it smaller. This establishes weak logical support across all the relevant distributions; the fair-die calculation alone would not.
3. The even outcomes greater than three are four and six. Neither is two, so Pr([TWO] | [EVEN] ∩ [HIGH]) = 0 / (2/6) = 0 < 1/6. Under this distribution, the enlarged premise set counts against the conclusion. This one distribution is also a counterexample to weak logical inductive validity, which requires non-decrease under every distribution.
4. The new premise rules out the conclusion while remaining compatible with the old premise. For another example, a card is drawn uniformly from cards numbered 1–4. Learning that its number is at most 2 raises the probability of 1 from 1/4 to 1/2. Learning in addition that it is even leaves only card 2, so the probability of 1 falls to zero.

Solution

1. For A, the increase is 0.8 − 0.5 = 0.3. For B, it is 36/37 − 0.9 = 27/370, approximately 0.073. Both likelihood ratios are 0.8 / 0.2 = 4.
2. Probability increase ranks A higher. Log-likelihood support is log(4) in both, so it ranks them equally. The measures capture different aspects of support.
3. No. We have checked material support under two particular functions. Logical support requires the condition to hold under every admissible probability function, keeping the propositions’ logical relationships fixed.
4. We need data and assumptions about the situation to estimate or justify a probability distribution. This is a statistical and subject-matter task, distinct from calculating support once a distribution is given.