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0

Think about $A$. There are only two cases. Either $A$ goes to the party or does not. Say he does. But $A$ goes only if $B$ does so consequently so does $B$. Say $A$ does not go to the party. Then $B$ won't either by the second statement. There are only two cases. Both party Both don't


1

Your definition of logic is pretty much correct. A logic contains both the language which the signature $\Sigma$ generates and the deductive system defined by $\vdash$. A type theory is a logic with different sorts of individuals (called "types") and constructions that generate new types form existing ones, like product and arrow types. An internal logic ...


0

This is a sequel of the selected answer in which @Mauro discovered the similarities between ✳113.44 and ✳166.44 and suggested that their dems are analogous to each other. $×α‘β$ is convenient because it is viewed as the descriptive function $×α$ taking $β$ as its input. By ✳38.11, $ ×α‘=α×β $ By ✳113.1, $ α×β=s‘α\underset{,,}{\downarrow}‘‘β $ By ✳37.101, ...


1

I looked through the comments briefly and nobody seems to have made the following point (if I missed it I apologize in advance). In physics there is a built-in "universe" of objects (or so they think) so when we speak of the mass of an object, the object in question ranges through this built-in range. In mathematics, the range is context-specific. Perhaps ...


-1

By what is given, you have to negate the following statement : It is raining if and only if the sun is shining or there are clouds in the sky. The solution is : It is not raining if and only if the sun is not shining and there are no clouds in the sky


0

I give a quite detailed answer at the linked question, that tries to explain in detail how each event changes the state of (mental) affairs on the island, and how much of "common knowledge" is really required to eventually cause the individuals concerned to leave. In my notation there with $C$ for "it is common knowledge that" and $E$ for "everybody knows ...


3

Yes, and no. Essentially, sometimes. Consider the function $f(x)=-x$ for $x\in\Bbb R$. Then what would $f^{(\omega)}(x)$ be? It should be a limit of $f^{(n)}(x)$. But there is no limit. Or consider the function $f(x)=x+1$ for $x\in\Bbb N$. Then $f^{(\omega)}(0)$ is what? But consider the function $f(x)=x+1$ for $x\in\omega_1$, then ...


1

If you want to formalize the notion that no one can be married to themselves, it would be best to introduce another statement (an axiom) in addition to the answer given (itself an axiom) something like: $$\forall x (\neg M(x,x))$$ You could also formalize other logical and legal requirements for $x$ and $y$ to be married (e.g. that they are both human ...


0

I'd say that you were correct and that the statement in question is false. The sequence of finite partial sums of such a series goes [-1, 0, -1, 0, ...]. Thus, if you pick any sufficiently large n for a partial sum, you will get a number which lies infinitely close to a member of the set (-1, 0) (two numbers are infinitely close if the absolute value of ...


2

First look up the definition of an infinite sum: $$\sum_{n=0}^{\infty} G(n) = \lim_{k\rightarrow \infty} \sum_{n=0}^{k} G(n)$$ Then look up the declaration of a limit: $$\forall \Delta_F > 0, \, \exists x_0 ,\, \forall x > x_0 : |F(x) - L| < \Delta_F {\color {red} {\iff \atop \rightarrow}} \lim_{x \rightarrow \infty} G(x) = L$$ And the question ...


1

Replacing that axiom would make certain structures into models that were originally not models. For example take standard $\mathbb{R}$ but define every number in the interval $[-10,-5)$ to be greater than every number in the interval $[-5,-4)$. (Additionally we must specify that $[-4,\infty)$ comes after $[-10,-5)$ and $(-\infty, -10)$ comes before ...


1

Yes, currently the problem is open whether or not the Partition Principle and the Axiom of Choice are equivalent. There are two major factors for this (in my opinion): Many people become less interested in choiceless results. So while they might be very happy to hear about them, they prefer to put their research efforts towards other directions. The ...


1

Any good mathematician knows that to decide whether someone is right or wrong you must first define a framework where you can be either right or wrong! So let's say the framework is basic set theory. Whether it's true or false depends on how we phrase things. I would say this: The statement $'' \sum_{n = 1}^{\infty} (-1)^n \; \text{is a real number}''$ is ...


0

I believe you are correct, and your professor is incorrect. The summation is a syntactically valid statement, and it has a canonical semantic meaning. However, a non-existent limit is not a real number in much the same way an unicorn is not a real number. It is certainly valid to ask, and people who know the definition of a real number will simply say no, it ...


1

A number of efforts have been made as listed below. Patrick Suppes provides an excellent overview and succinct survey of the issue of division by zero in his Introduction to Logic, Chapter 8.5, The Problem of Division by Zero, and Chapter 8.7, Five Approaches to Division by Zero. http://stuff.mit.edu/afs/athena.... Suppes notes that the fourth of the five ...


2

Sure. Lots of math programming libraries essentially do this by adding NAN (not a number) as a possible value of functions. Things don't break because NAN is not the domain of any function/operator/etc. However, saying anything divided by zero equals 42 would result in an inconsistent theory, which makes everything true. We see lots of examples of this in ...


3

In some contexts, things like $5/0$ should be taken to be $\infty$, where that is neither $+\infty$ nor $-\infty$, but a single $\infty$ at both ends of the real line or at the extreme of the complex plane. This makes all trigonometric functions and rational functions continuous. This is really an example of one of the points at infinity in projective ...


1

The proposed alternative, $\forall x\forall y\left(\left(M(x,y)\implies F(x,y)\right)\land (x\not=y)\right)$, is simply a false statement: Examined carefully, it implies $\forall x\forall y (x\not=y)$, which patently fails when $x=y$. An alternative that might be arguably correct is $$\forall x\forall y\left(\left((x\not=y)\land M(x,y) \right)\implies ...


1

I would say, no. The statement $\forall x \forall y (x \ne y)$ is false for any nonempty domain of discourse. If you want to add an "acceptable couple" predicate $A$ (in your case $A(x,y)$ might be defined as $x \ne y$) it would look something like $\forall x \forall y ( A(x,y) \rightarrow (M(x,y) \rightarrow F(x,y)))$. This says "if the two are an ...


1

If one CAN be married to oneself, $M(x,x)$ will be true for that person.


1

Without more details, it is difficult to say what sort of operations you are allowed to use. Here is one possible way of doing this by using unions: $Q(n + 1) = Q(n) \cup \{X \cup \{n + 1\} : X \in Q(n)\} \cup \{\{n + 1\}\}$


1

I'm still not sure how you are asked to answer the problem, according to your Lecture Notes ... 1) use the equivalence $p→q$ with $¬p∨q$ to rewrite it as : ¬P(x)∨Q(x)∨M(x) 2) after the correction : ¬Q(x)∨R(y,x)∨¬P(y) 3) apply the above equivalence followed by Double Negation to get : M(y)∨¬(¬P(x)∧R(x,y)), and then apply DeMorgan [$¬(p∧q)$ is equivalent to ...


0

The blue-eyed people determine their eye colour by a proof-by-contradiction that creates hypothetical people each of whom uses a proof-by-contradiction based on hypothetical people etc. It assumes that every one of these hypothetical people is able to fully reason out the thinking of each of the hypothetical people they think of. In order for the proof to ...


1

This is just an illustration for 113.44: Suppose $k=\{ \beta, \gamma, \delta, ... \}$ Then $(\times \alpha)‘‘k$ $ = \{\times \alpha ‘\beta, \times \alpha‘\gamma, \times \alpha‘\delta, ...\} $ $=\{\beta \times \alpha, \gamma \times \alpha, \delta \times \alpha, ...\}$ $=\{s‘\alpha \underset{,,}{\downarrow}‘‘\beta, s‘\alpha ...


2

Your professor certainly isn't right that "no fathomable meaning" can be assigned to the expression $\sum_{n=1}^{\infty}(-1)^n$. Otherwise, what is meant by the following statement? The series $\sum_{n=1}^{\infty}(-1)^n$ is divergent. But in practice one frequently conflates the description of an infinite series with the limit of its partial sums. ...


-1

I'd say, you're right since one can define a set of arcane and strange numbers, in which this series convergates to a certain element of the set. Since this number is not real, the proposition is false.


0

If $S^;P$ is like $S‘‘C‘P$ in an order similar to $P$, then in picture: $S‘R$-----$S^;P$------$S‘T$ $S\uparrow$------------------$\overset{\smile}{S} \downarrow$ $R$----------$P$---------$T$ Given $R=Q‘T$,When the correlator is $ \times P$, then $\times P^;Q$ can be pictured like this (from bottom up): ...


4

As indicated in the other answer, there are countably many undecidable statements (the largest possible number), and this is an immediate consequence of the incompleteness theorem. The version of incompleteness that Andrews is using is that if $T$ is a consistent, recursive set of axioms that interprets a modicum of arithmetic, then the set of $T$-decidable ...


4

Assuming consistency, there are infinitely many undecidable statements. The set of decidable statements is recursively enumerable, but not recursive. Which means the set of undecidable statements is not recursively enumerable - that is, any computer program which enumerates only undecidable statements will fail to enumerate all undecidable statements. So ...


-1

I think dismissing an argument that stands foursquare in a 900 year tradition of logical discourse as a 'strange apparent aberration' is a little questionable. Anyway, it's worth noting that the argument's overly strong axioms and definitions lead to modal collapse. Modifications have been suggested that appear to solve this issue. For example see ...


0

The answer is yes. Look for Lagrange Interpolation polynomial. You are looking for a function $f$ such that $f(1),..,f(26)$ takes some particular values. The Lagrange interpolation sais that no matter what those values are, the following polynomial works: $$f(1) \frac{(x-2)(x-3)...(x-26)}{(1-2)(1-3)...(1-26)}+f(2) ...


2

$$\begin{align} q\iff (\lnot p \lor \lnot q) &\equiv [q \rightarrow (\lnot p \lor \lnot q)] \land [(\lnot p \lor \lnot q ) \rightarrow q]\\ \\ &\equiv (\lnot q \lor \lnot p\lor \lnot q) \land [\lnot(p \land q) \rightarrow q)]\\ \\ & \equiv (\lnot q \lor \lnot p) \land [(p \land q) \lor q]\\ \\ &\equiv \lnot (q \land p) \land [(p \land q) \lor ...


5

It is very hard to "disentangle" this formula. Start with the LHS of $✳166.44$ : $\Sigma‘ \times P^{;} Q$. The first step is to apply the "transformation" $✳166.1$ : $Q \times P = \Sigma‘P \underset{\overset{\textbf{.,}}{\ }}{\downarrow}^{;} Q$. Note In the following steps I'll use "$\downarrow$" in place of "down-arrow with dot & comma". ...


0

It's not an equivalence, it's an implication. We need to show: $(A\wedge B)\vee(\neg B\wedge C)\to (A\vee C)$ $\begin{align} & (A\wedge B)\vee(\neg B\wedge C) & \text{Premise} \\ \iff & ((A\wedge B)\vee\neg B)\wedge ((A\wedge B)\vee C) & \text{Distribution} \\ \iff & ((A\vee\neg B)\wedge (B\vee\neg B))\wedge ((A\wedge B)\vee C) & ...


1

$\neg\exists x\;\exists y\;P(x,y)$ is parsed as $\neg(\exists x(\exists y(P(x,y))))$ Or in English: "There does not exists two people who are parent and child." However, I know of several such people. Remember also that the negation of an existential quantifier, is the universal quantifier of a negation. $$\neg\exists x\;(Q(x)) \qquad\equiv\qquad ...


1

$ \exists x \exists y P(x, y) $ would mean "there exists $x$ ( for which ) there exists $y$ such that $x$ is a parent of $y$". Now the negation at the beginning I would think is applicable to the first quantifier. So we may as well bracket it $(\lnot \exists x)$. This means there does not exists $x$. Replace this in our first interpretation and we get, ...


5

$\exists x [P(x) \wedge \forall y [P(y) \to y\leq x]$ This is read from left to right as: "There exists a number such that it is perfect and for all numbers that if they are perfect it follows that they are less than or equal to it." In more natural language, this can be stated as: "There exists a perfect number that will be greater than or equal to every ...


2

You messed up the order of the quantifiers. What you wrote translates back to be: $$ \forall y ~ [P(y) \to \exists x ~ (P(x) \land y \leq x)] $$ In your version, this perfect number $x$ might depend on the perfect number $y$ that is chosen. In the original version however, the perfect number $x$ must be chosen first (independently of $y$). To illustrate ...


5

The problem with your translation is that there is ambiguity about the scope of the quantifiers you use. In particular the sentence expresses the claim that "There is a perfect number, so that every perfect number is less than or equal to the first." Or as Git Gud says, there is a maximum perfect number. Unfortunately, what you have written is, most ...


1

I've read the whole problem in Shoenfield's book, and I think it is quite clear that the author is guiding the readers to the concept of expressively adequate set of connectives (those that can express all truth-functions of the language), splitting the process in several steps. And I am not sure what the author means by "certain of the $H_{d,n}$ and ...


0

Hint: Write out all possible combinations for truth values of $A,B,C$. There are eight. Evaluate $A \bigvee B$ for all eight. Evaluate $\neg B \bigvee C$ for all eight. Evaluate the and of the two above expressions for all eight. Evaluate $A \bigvee C$ for all eight. Compare.


1

Sounds correct to me. You're essentially proving this by induction on $n$. And, yes, it's kind of ugly because you have to go through the 16 cases once, but I think there's no "elegant" way around that.


2

You always have $B\lor \lnot B$. If $B$ then $\lnot B\lor C$ implies $C$ which on its turn implies $A\lor C$. If $\lnot B$ then $A\lor B$ implies $A$ which on its turn implies $A\lor C$.


6

$$(A\lor B) \land (\lnot B \lor C)$$ If the premise above is true, then by conjunction elimination, $A\lor B$ is true $(1)$ and $\lnot B \lor C \equiv B\rightarrow C$ is true.$(2) $(1) \;A:\;$Suppose A is true. Then $A\lor C$ is true ($\lor$-Introduction). $(1)\;B:\;$ Suppose B is true. Then by modus ponens with $(2): B\rightarrow C$, we have that $C$ ...


1

I doubt there is any such argument. Moreover people learn best by small examples which introduce one idea at a time. Certainly not by big examples with many new things. The way logic reasoning is taught is by introducing one logical operator at a time then showing some examples, next solving exercises and then checking if they were solved correctly. This is ...


0

If $n=1$ and no one knows any more than they can see, then the one blue-eyed person does not know that there are blue-eyed people on the island. If $n = 2$, then blue-eyed person $a$ does not know that blue-eyed person $b$ knows that there are blue-eyed people on the island. For all $a$ knows, $b$ could be alone in his blue-eyed-ness, in which case $b$ ...


2

Use conjunction elimination, or take the long route. $\left.\begin{align} (p\wedge q) \to q & & \text{premise} \\ (\neg p \vee \neg q) \vee q & &\text{conditional} \\ \neg p \vee (\neg q\vee q) & &\text{disjunctive associativity} \\ \neg p \vee \top & &\text{disjunctive negation} \\ \top & &\text{universal bound} ...


4

Since you have the antecedent $[(\sim p \vee q) \wedge p ]$ reduced to $q \land p$ (which is correct), you simplify (or you might call it $\land$-elimination) to get $q$, as desired. That is, $$q\land p$$ $$\therefore q$$ Hence, we can claim that $[(\sim p \vee q) \wedge p ] \Rightarrow q$. (Therefore $p$ follows as well, but you are asked to show ...


1

I know that both NF and NFU are supposed to be quite weak. NFU is weaker than PA and it has the same consistency strength as MacLane set theory (Z with bounded quantifiers). MacLane set theory is significantly stronger than PA. It has the same consistency strength as Russellian unramified typed set theory (TST). In an answer to another question, I tried ...


0

I'm not sure how you intend to get to a disproof from your observation. This has an easy solution via the compactness and completeness theorems, though. Let $T$ be the any first-order theory for which every structure in the class models $T$. For every chain of inequalities of finite length $$ a < fa < f^2a < \ldots < f^n a $$ we can find a ...



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