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1

Hint: An implication is false precisely when the hypothesis is true and the conclusion is false.


0

Option #1: $\neg(a\leq15\implies{a>1})\iff$ $\neg(\neg(a\leq15)\vee(a>1))\iff$ $(a\leq15)\wedge\neg(a>1)\iff$ $(a\leq15)\wedge(a\leq1)\implies{a\leq1}$ Option #2: $\neg(a\leq15\implies{a>1})\iff$ $\neg(\neg(a\leq15)\vee(a>1))\iff$ $\neg((a>15)\vee(a>1))\implies$ $\neg(a>1)\iff{a\leq1}$


0

The branch of science is called Decision Theory/Science, Management Science or Operations Research. Some of the tools that are useful Influence Diagrams http://www.cs.ru.nl/~marinav/Teaching/BDMinAI/influencediagrams05.pdf Decision Trees (Issue Trees are also related) http://en.wikipedia.org/wiki/Decision_tree Causal Loop Diagrams


0

Your reasoning looks fine to me. For comparison, here is a calculational approach which may shed some additional light. $ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\then}{\mathrel{\Rightarrow}} ...


0

The earlier answers all pull $\;y=1\;$ as a rabbit out of a hat. Let me show you a way to avoid that. $ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\then}{\mathop{\;\Rightarrow\;}} ...


3

Take $y = 1$ and show that $x \neq x \implies x = 0\tag{1}$ Note that you can express this in its logically equivalent contrapositive form, which might help you see that the statement is in fact true. Take $y = 1$. Then we have $x\neq 0 \implies x = x$, which is certainly true, because $x = x$ is always true. So no matter whether $x = 0$ or $x\neq 0$, the ...


0

One way to examine whether a "there exists" statement is trus is to see whether it's true for some particular values. Suppose you say "If I pick $y = 1$, is the rest true?" You might have a little trouble answering that, so move on. Try saying "If I pick $y = 2$, is the remainder true?" What result do you get?


4

It is true. $y=1\in\mathbb R$ does the job. Note that $x\neq x\Rightarrow x=0$ is actually the same statement as $x=x\vee x=0$. This statement is evidently true for each $x\in\mathbb R$.


0

You have to take into account that already Aristotle discussed "chains" of predication, i.e. chains of syllogisms (and he considered also the infinite case); see : Aristotle, Posterior Analytics (editor Jonathan Barnes, 2nd ed, 1993), A21-22, page 29-on and the detailed discussion in : Jonathan Lear, Aristotle and Logical Theory (1980), page 17-on.


3

The only thing about For all real numbers $x$ with $x^2-3 x+2\leq 0$, $1\leq x\leq 2$ that seems to need improvement is that having just a comma between the formulas makes it difficult at a glance to see where one formula ends and the next begins. The standard solution for that is to insert a spacer phrase to get something like For all real numbers ...


1

How about Lewis Caroll, in particular his book Symbolic Logic example What conclusion may be drawn from: (a) No interesting poems are unpopular among people of real taste. (b) No modern poetry is free from affectation. (c) All your poems are on the subject of soap-bubbles. (d) No affected poetry is popular among people of real taste. ...


5

There's only one variable, so there should only be one quantifier. I would go with something like: $$ \forall x \in \mathbb R \text{, } ~~ [x^2 - 3x + 2 \leq 0 \implies 1 \leq x \leq 2] $$


1

For millennia, syllogistic logic as described by Aristotle was widely considered to be the fundament of all logic. It was only in the late 1800s with the work of Frege and his successors that it was recognized that the Aristotelian structure is not actually strong enough to represent many kinds of reasoning that are used routinely in mathematics. What ...


0

If you allow for a board with an infinitely repeating looping section (to emulate the infinite tape), I'm pretty sure you can develop a turing complete chess board. Pawns probably can't be used (because they can't be moved back to loop a calculation) and queens probably have too much movement range to be useful as well. You'll probably need "walls" on your ...


1

If we multiply $X + Y \ge \frac T{a+b}$ by $a+b$ we get $aX + bX + aY + bY \ge T$ This will be more likely to be true than $aX+bY \ge T$ because we are adding positive terms to the left. So in your question the sign should be $\lt$


0

First of al in logic $ B \times C$ is just an other set. so lets assume Ax is true iff $ x \in A$ Bx is true iff $ x \in B \times C$ Rxy is true iff $ f(x) = y $ then your relations become $f(x)$ is a function: $ \forall x (Ax \to ( \exists y (By \land Rxy \land \forall z ((Bz \land Rxz) \to y = z )))) $ $f(x)$ is injective: $ \forall x y z ((Ax ...


-1

First, I would like you tho think about the relation between $\cup$ and $\land$, and $\cap$ and $\lor.$ Then, prove this statements using true and false propositional definitions. Once done, you will know how to prove this problem on your own.


2

Consider $P,Q,R=1$. The first statement is true and the second statement is...


1

It depends a bit how you count / distinguish sets of sentences, If you assume that sets are closed under deduction then there is only one inconsistent set (the set of all sentences) all other sentences are consistent so the chance that you have the inconsistent set is almost nil. If you don't assume closure it is the opposite, every consistent set has ...


0

As I commented, this is false for infinite sets. To understand why, I think it is simplest to forget about elements of $T$ that are not in the range of $R$ and then to forget about the distinction between different elements of $S$ that have the same image under $R$ in $T$. When you have done that, you can also forget about induction. In more detail: The ...


3

What you have there is a Disjunctive syllogism. It is named after the logical disjunction (or the or operator, $\vee$). The formal name of the is modus tollendo ponens, and is an example of a valid syllogism, and is hence also a sound syllogism, provided it is a correct dichotomy of the two ideas, and that both have a true value. Example Here is the formal ...


0

You need a binary relation $A$; the intended interpretation of $A(x,y)$ is that regions $x$ and $y$ are adjacent. You need a constant for each region, so that you can specify in your theory which regions are adjacent. You need a way to specify a $k$-coloring. There are several ways to do this; one is to have unary relations $C_1,\ldots,C_k$ and sentences ...


3

As you observe, $p\oplus q\oplus r$ is true when only one of the variables is true, or when all three are true. Therefore, $(p\oplus q\oplus r) \wedge \neg(p\wedge q\wedge r)$ would fit your needs.


3

The simplest formula is obviously, $$ ( p \land \lnot q\land \lnot r) \lor ( \lnot p \land \lnot q\land r) \lor ( \lnot p \land q\land \lnot r) $$ Maybe there is a more succinct and elegant formula but I can't seem to unearth it.


1

For syntax and semantics of higher-type classical logic, you can see : Part I : CLASSICAL LOGIC, of Melvin Fitting, Types, Tableaus, and Gödel’s God (2002).


1

I'm able to prove it "independently" from the Deduction Theorem, but the proof is quite longer ... The axioms are : $F \rightarrow (G \rightarrow F)$ $(F \rightarrow (G \rightarrow H))\rightarrow ((F \rightarrow G) \rightarrow (F \rightarrow H))$ $(\neg G \rightarrow \neg F) \rightarrow ((\neg G \rightarrow F) \rightarrow G)$ For ...


1

You're going backwards. It's not necessarily true that if $\mathfrak{A}\equiv \mathfrak{B}$ then $\mathfrak{A}\cong\mathfrak{B}$: for instance, the field of algebraic numbers is elementarily equivalent to the field of complex numbers, but simply by cardinality they're not isomorphic.


0

In first order logic, there are validities in addition to tautologies. These statements are statements that are true in every possible interpretation, and just not being a tautology isn't enough to tell you if it's a valid statement or not. Gödel's incompleteness theorems suggest that for every system that's powerful enough to represent arithmetic, there are ...


2

In oder to prove 5 and 6, we need some preliminary results. T1 : $P \rightarrow P$ 1) $P \rightarrow ((Q \rightarrow P) \rightarrow P)$ --- Ax.1 2) $P \rightarrow (Q \rightarrow P)$ --- Ax.1 3) $(1) \rightarrow ((2) \rightarrow (P \rightarrow P))$ --- Ax.2 4) $P \rightarrow P$ --- from 3), 1) and 2) by Modus Ponens twice. T2 : $(Q \rightarrow ...


3

For the non-trivial direction, you're given a model $B$ of the positive consequences of $\Gamma$ and you want an $A$ satisfying two requirements: (1) $A\subseteq B$ and (2) $A$ is a model of $\Gamma$. Notice that requirement (1) can also be phrased as "$A$ is a model of $\Delta$" for a suitable $\Delta$, namely the set of negations of all the sentence ...


1

Formally it doesn't "come from" anywhere -- it is a definition. Intuitively, the free variables in a formula are all the variables that appear in it, except that it doesn't count when variable appears within the scope of a quantifier that binds. The symbolic definition simply makes precise what that intuition means. The case you quote is a shorthand for ...


3

Here's one way to do it (I think). Starting with $B$, we let $B'\subseteq B$ be a minimal model of the positive consequences of $\Gamma$. $B'$ can be obtained by transfinite recursion (taking intersections at limits). Now, we will show that $B'\vDash \Gamma$. Suppose not; that is, suppose $B'\vDash \neg \phi$ where $\phi$ is a consequence of $\Gamma$. It ...


-3

OTTER has found an 8 condensed detachment, level 5 proof: ----> UNIT CONFLICT at 11.16 sec ----> 26226 [binary,26225.1,2.1] $F. Length of proof is 8. Level of proof is 5. ---------------- PROOF ---------------- 1 [] -P(i(x,y))| -P(x)|P(y). 2 [] -P(i(i(n(a),n(b)),i(b,a))). 3 [] P(i(x,i(y,x))). 4 [] P(i(i(x,i(y,z)),i(i(x,y),i(x,z)))). 5 [] ...


1

The proof must be : 1) $\lnot p \land \lnot q$ --- premise 2) $\lnot p$ --- form 1) by $\land$-elim 3) $\lnot q$ --- form 1) by $\land$-elim 4) $p$ --- assumed [a] 5) $\bot$ --- from 2) and 4) by $\lnot$-elim 6) $q$ --- from 3) and 5) by RAA (or Double negation) 7) $p \rightarrow q$ --- from 4) and 6) by $\rightarrow$-intro, discharging [a] 8) $q$ ...


1

No, that is not correct. I'm assuming it's supposed to be some kind of natural deduction system, but the deductions you annotate with $\neg$elim and $\to$into don't follow any sane negation elimination or implication introduction rules I know. For example you try to conclude $p$ from $\neg p$. That makes no logical sense "Socrates is mortal, ergo Socrates ...


3

I've found a related result into : Jon Barwise (editor), Handbook of mathematical logic (1999), A.2 : H.Jerome Keisler, Fundamentals of Model Theory, page 47-on. See page 72 : 5.10 : Lyndon Homomorphism Theorem (Lyndon [1959]). We can derive from the proof of it the application to the propositional case. Let $\Gamma$ consistent, and let $\Gamma^+$ the ...


1

You can see : James Garson, Modal Logic for Philosophers (2006), Ch.10 : Axioms and Their Corresponding Conditions on $R$, page 209-on. we will prove a theorem that may be used to determine conditions on $R$ from axioms (and vice versa) for a wide range of axioms (Lemmon and Scott, 1977). (For a more general result of this kind see Sahlqvist, 1975 ...


1

The standard abbreviation for "there are at least $482$ objects $x$ satisfying $\phi(x)$" is $$ (\exists^{482} x)\phi(x) $$ The extra quantifiers of the form $(\exists^{n} x)$ can either be defined directly in the metatheory, or viewed as abbreviations for longer formulas. So you could write $$ (\exists^{482} x)\phi(x) \land \lnot(\exists^{483} x)\phi(x). ...


1

$$\exists x_1, \ldots, x_{482}\left(\bigwedge_{i \neq j \leq 482}x_i \neq x_j \wedge \forall y \left(\phi(y) \leftrightarrow (\bigvee_{i \leq 482} y = x_i)\right)\right)$$ Or, since it's not in doubt that there is such a first-order formula, create some ad-hoc shorthand and write "$\exists^{482} x\, \phi(x)$ where $\exists^{482}x$ is a quantifier denoting ...


1

Let $S_P$ be the set of individuals characterized by $P$ and $S_Q$ the set of individuals characterized by $Q$, $\mid S_P \cap S_Q\mid=482$. Where $\mid S \mid$ is the cardinality of the set $S$. You can possibly define these sets as $S_P = \{x \mid P(x)\}$.


0

Yes, there is an unrefutable argument that $2^ω$ is not countable, namely Cantor’s diagonal argument mentioned by @user72694. What exactly this argument says? Apart of the rest of set theory, as well as confusions between a theory and metamathematics; see @Hurkyl’s answer for analysis. Cantor’s diagonal argument shows that a surjective mapping from $\mathbb ...


2

First of all, the word "group" has a technical meaning which is not the meaning you're using here. It seems to me that you want something more like a "theory." A "theory," in the technical sense, is essentially a collection of statements in some logical language. The subjects that study logical theories are, from two different directions, proof theory and ...


2

$T_4$ is not conservative over $T_3$ but $T_5$ is. For the first claim, consider the following statement in $L_2$: $\chi \equiv \forall x, y(x \cdot y = x)$ Suppose $\chi$ were true. From $\xi$ it would follow that $\forall x(x = x + x)$. Now, consider an arbitrary $x$. By $\psi$ there is some $n$ such that $x + n = x$. By $\eta$ there is some $y$ such ...


0

Make a drawing! Assume that $P,P'$ are two planes. $t_{u_0}(P)=P'$ iff {$P,P'$ are parallel and there are $B\in P,A\in P'$ s.t. $u_0=\vec{BA}$}. Note that $u_0=\vec{BA}$ iff $t_{u_0}(B)=A$. Finally the first assertion implies the second one ; yet the converse is false.


0

I would formalize the proposition "I know that I don't know you" in the following way: First I would introduce the relation $K(x)$ via $$K(x) :\Leftrightarrow \text{I know that x}$$ whereby $x$ is a proposition. Let $A$ be the proposition "I don't know you". So your given proposition "I know that I don't know you" is $K(A)$. The negation of $K(A)$ is $\neg ...


0

If possible suppose $B\subseteq C$.AS $A\subseteq B$,that would imply $A\subseteq C$,which is a contradiction.


1

A proof of the problem may be obtained if you assume $A\subseteq B \land B\subseteq C\land A\not\subseteq C$. Because, $$(A\subseteq B \land B\subseteq C)\land A\not\subseteq C\implies A\subseteq C \land A\not\subseteq C$$ Which surely is a contradiction.


4

The flaw in your solution is "from $x\in A$ and $A\not\subseteq C$, it follows that $x\not\in C$." This may not be true for every $x\in A$, but the hypothesis says that this is true for at least one. This is why the solution that you are given uses existential quantification. For a concrete example, consider $A=\{0,1\}, B=\{0,1,2\}, C=\{0\}$. It is not true ...


0

Intuitively they are actually not the same. $T(x) ⇒ \neg S(x)$ means "( if $x$ is a TA, then $x$ has not taken the class )", which says nothing about whether $x$ is a TA or not.


2

Yup, you're right OP- all these are all correct. As servabat pointed out above, C is true because all x for which ${x^2<x}$ is true are between 0 and 1, so they satisfy the condition of being less than 1. The reason for this is that C doesn't claim "All ${x<1}$ are a solution to ${x^2<x}$". Instead, it means "All solutions to ${x^2<x}$ are less ...



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