15

"Some" does not exclude "all", but you cannot deduce "all" from "some". Having said that, the above argument is not valid. From premises 1 and 2 we can derive : Some reasonable people are criminal that is equivalent to : Not all reasonable people are not criminal. Having said that, from "Some reasonable people are criminal" we cannot conclude by logic ...


4

A statement p is more general than statement q if p pertains to more instances than q (i.e. when all the instances of q are instances of p, but not vice versa). A statement p is universal if it does not contain any references to particular objects, times, places etc. And so Never shoot anybody who wears a hat and yellow socks is universal, because it ...


4

One way to interpret the order of nested quantifiers is as expressing dependence relations among choices of objects selected to satisfy the quantified formula. Friedman in Kant's Theory of Geometry explains how the inability to express such dependencies in syllogistic (you need three nested quantifiers to define limits, for example) forced early calculus and ...


4

You are trying to prove: ∃x (Fx → ∀x Fx) and your derivation starts with the assumption Fx. If so, the step with the Universal Generalization, deriving ∀xFx from Fx, is wrong. The rule has the proviso that the variable to be quantified, in this case x, must not be free in any assumption; in the proof, we have Fx as assumption, and x is free in it. The ...


4

Your rephrasings of formulas in words are correct, but in this case moving the quantifier makes no logical difference (classically). You can verify this by converting formulas into equivalent form without implications using A → B = ¬A ∨ B, and then using the fact that quantifiers can be freely moved across conjunction and disjunction as long as the variables ...


4

The modern term is due to George Boole, with his booklet of 1847: The Mathematical Analysis of Logic, aiming at the application of mathematical tools and methods to the study of logic. For a recent historical overview, see: Leila Haaparanta (editor), The Development of Modern Logic (2009). In the same booklet, Boole set also a link with language: [...


4

The negation of "there are at least two..." is "there is at most one..." i.e. either "there is no..." or "there is one and only one...". Therefore ¬❷xPx ≡ (¬∃xPx) ∨ (∃!xPx) where ∃! is the unique existential quantifier defined by ∃!xPx ≡ ∃*x(Px* ∧ ∀y(Py→y=x)) By definition ✌x¬Px ≡ ¬❷xPx Hence (replacing Px by ¬Px) ✌xPx ≡ ¬❷x¬Px ≡ (¬∃x¬Px) ∨ (∃!...


4

HINT: I'll sketch the derivation. Since the theorem is a conditional, try using conditional proof/conditional-introduction by assuming P(a) and trying to derive ∀x(P(x) ∨ ¬(x = a)) from it. Here, to derive it, I would try an indirect proof by assuming the negation ¬∀x(P(x) ∨ ¬(x = a)) and trying to derive a contradiction. Use quantifier equivalence rules to ...


4

Some newspaper readers are criminal This means that at least one newspaper reader is a criminal. It can be more, it can even be all of them. We do not know. But at least one is. All newspaper readers are reasonable people All of the newspaper readers are reasonable people. Because at least one of the readers was a criminal, we must have one reasonable ...


4

Yes - the key term is "generalized quantifiers." They are studied in the contexts of both natural language and in mathematical logic. I'll focus on the logic side, about which I know more. A name which crops up in both contexts is Jon Barwise, and this article of Vaanaanen describes much of Barwise's work on generalized quantifiers; this paper of Barwise on ...


3

It says that (∀x)Fx ⇒ Fr where F(x) is any sentential formula in which x occurs free, r is a term, F(r) is the result of substituting r for the free occurrences of x in sentential formula F, and all occurrences of all variables in r are free in F. Note that that way of putting it doesn't say that x is free in (∀x)F(x). It says that x is free in F(x) — in ...


3

No. On what I take to be the most natural interpretation, two logically-equivalent ways of rendering the first statement in formal logic would be: For any houses/each house in the area, it's not the case that food can be found in them There does not exist a house such that food can be found in it. Neither sentence entails that a house exists. However, ...


3

Start with "intuition" : if there is a black cat, then there is a cat is an argument that "sound" quite correct. To prove it "semantically" we have to assume an interpretation I in which the premise is true. This means that there is an object o of the domain of the interpretation I for which both F and G holds. But this implies that for that object o ...


3

Resolved! I realised that I was going nowhere by assuming the opposite of what I was given as premise... I obviously had to assume the opposite of what I was trying to prove:


2

This was a well-known problem in the middle ages. They solved it by distinguishing the existential import of affirmative from negative propositions. An affirmative proposition does have existential import. If we read "there is something that does not exist" as (1) something is non-existent then it is false, because it is affirmative, and thus implies the ...


2

Yes, you can you correctly infer from the ordinary language sentence "Food cannot be found in any house (in this area)" that "There is AT LEAST [one] house (in this area)". This is because the existence of houses in the area is presupposed, not asserted. Note that the first reply confuses the way the English sentences would be rendered formally, with the ...


2

Well, it's late at night so I'm sure I'm going to repeat past mistakes and screw something up here. But let's give it a try. You start with the original sentence: ∀w(∀v((v=w∧φ(v))⇔φ(w))) From there we just need to discharge the outermost quantifier, instantiating to the term "t" (assuming we have such a term laying around): ∀v((v=t∧φ(v))⇔φ(t))) Then ...


2

Let me ask the flip question: even if I measure something, why should it be correct? Just because something is measurable doesn't mean that it is not a delusion. Astrology uses extensive measurement but its not really true. Eugenics used measurement but it wasn't really true knowledge. There seems to be a fundamental fallacy of presumption in this question. ...


2

In social science and humanities, qualitative research methods systematically collect observations, identify phenomena, and analyze them without quantification. Consider this report by the research nonprofit FrameWorks. (There are plenty of peer-reviewed examples — qualitative methods are widely used in anthropology, for example — but this was the first ...


2

It depends on the rules you are allowed to use... Here is a proof using Double Negation elimination : 1) ¬ (Fx ∨ ¬Fx) --- assumed [a] 2) Fx --- assumed [b] 3) Fx ∨ ¬Fx --- from 2) by ∨-intro 4) ⊥ --- from 1) and 3) 5) ¬Fx --- from 2) and 4) by ¬-intro, discharging [b] 6) Fx ∨ ¬Fx --- from 5) by ∨-intro 7) ⊥ --- from 1) and 6) 8) Fx ∨ ¬Fx --- from 1)...


2

The statement "some are" means "there exists at least one". The statement "there exists at least one" is consistent with the statement "all are". This consistency between Some and All has to be accounted for. Thus the I statement, "Some S are P" becomes: "The ratio of subject S, in relation to predicate P, (1) is larger than zero percent, but (2) is less ...


2

These all appear to be correct.


2

1) ¬∀x(P(x) → Q(x)) --- premise 2) ¬∃xP(x) --- assumed [a] 3) P(x) --- assumed [b] 4) ∃xP(x) --- from 3) by ∃-intro 5) ⊥ --- contradiction : from 2) and 4) 6) Q(x) --- from 5) by ⊥-elim 7) P(x) → Q(x) --- from 3) and 6) by →-intro, discharging [b] 8) ∀x(P(x) → Q(x)) --- from 7) by ∀-intro 9) ⊥ --- contradiction : from 1) and 8) 10) ∃xP(x) --- from ...


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