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0

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'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

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 ...


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

Before you learn any shortcuts for solving this problem, you must learn to solve it the slow way. Write out your constraints and test all of the cases. $$\begin{align} F(A, B, C, D) & = (A \iff \lnot B) \\ & \land (B \iff C) \\ & \land (C \iff A) \\ & \land (D \iff (\lnot A \land \lnot B \land \lnot C)) \\ & \land (B \iff (\lnot A ...


1

I believe that the question has some mistake. If we represent person X telling the truth with boolean proposition x, then we can convert all five statements to a = ~ b b = c c = a d = ~a /\ ~b /\ ~c b = ~c /\ ~a From 1, 2, and 3. we get b = ~b which is a contradiction.


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 ...


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) & ...


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.


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

The proposition we want to prove is: "$5n + 3$ even $\implies$ $n$ odd". This is equivalent to the contrapositive statement, "$n$ not odd $\implies$ $5n + 3$ not even". Since an integer is either odd or even, it is equivalent to saying "$n$ even $\implies$ $5n + 3$ odd. If a number is even, then any integer multiple of it must be even. That is to say, $n$ ...


0

An even number is divisible by $2$, so we can describe an even number as $2k$. Then \[ 5n+3=2k \] \[ 5n=2k-3 \] \[ n=\frac{2k-3}{5} \] $2k$ is even so $2k-3$ is odd and an odd number divided by $5$ is still an odd number. Hence, when $5n+3$ is an even number, $n$ is an odd number.


1

Every integer is either even or odd. An even integer is an integer divisible by 2. If $n$ is even then $5n$ is also even, for it is divisible by 2. But since 3 is not divisible by 2, so that $5n+3$ is not divisible by 2, and hence $5n+3$ is an odd integer, qed.


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

Perhaps [1] could be of use. However, see [2] if you want a very abstract approach (an approach that is related to topics in the links I gave in my comment here). [1] Handbook of Mathematical Induction by David S. Gunderson (over 900 pages) [2] Elementary Induction on Abstract Structures by Yiannis N. Moschovakis


0

Hints: $$ (p \lor q \lor r \lor s) \equiv (((p \lor q) \lor r) \lor s) \; \text {Associative Law} $$ $$ (a \lor b) \equiv (\lnot (\lnot(a \lor b) ))$$ $$ (\lnot (a \lor b)) \equiv ((\lnot a) \land (\lnot b)) \; \text {de Morgan's Law}$$


1

Let $C$ denote the set of members of calculus class and let $D$ the set of members of discrete math class. The following statements are equivalent (explore step by step) and the last one is the Velleman answer: 1) Nobody in the calculus class is smarter than everybody in the discrete math class 2) For every person $x$ in $C$ there is a person $y$ in $D$ ...


4

Your answer asserts that there does not exist anyone $x$, who, iF $x$ is in Calculus, then (all students y are both in Discrete math and x is smarter than them.) This is clearly not what is conveyed in the original statement. What we need, essentially, is "There does not exist someone $x$ who is enrolled in Calculus AND such that, for all students y, if y ...


1

If $\forall x, x\ne m\Rightarrow L(x,m)$. The only person who doesn't like Mary is Mary herself. So if someone doesn't like Mary, that someone must be Mary. Or the contrapositive; if someone isn't Mary, that someone likes Mary. In your formulation, you have Mary liking Mary ($\forall x L(x,m)$) and Mary not liking Mary ($\neg L(m,m)$). These two ...


1

An inference (or argument) from given premises to a conclusion is valid if and only if there is no possible situation in which the premises are true and the conclusion is false. So, talking about propositional logic and truth-tables, the inference is valid (or the premises entail the conclusion) if and only if there is no valuation of the atoms (p, q, r,...) ...


4

Modus tollens does indeed apply. In each scenario, you are given two premises, and asked if you can infer, from the two given premises, the conclusion, $\lnot p$. It is also the case that we need to determine whether from $\lnot q$ we can then infer $\lnot(p\land q)$ in the first case, or $\lnot(p \lor q)$ in the second. In the first case, put $t = ...


0

From the other answers, the most convincing and reasonable explanation why logic implication is defined the way it is, is the idea of sufficient (versus "necessary") condition for something to be true. NOTE: I don't buy the answer/argument that "if-then" is not an equivalent definition of "=>"; it is just that we tend to have a different notion of "if-then" ...


6

It is true that if $A\cap B = \varnothing$, then $$(A\cup B) \setminus B = A$$ But we cannot simply assume that $A\cap B$ is empty. That is, $A$ and $B$ need not be disjoint. Informally, if A and B overlap, then removing all of $B$ from $A\cup B$ removes part of A, too. In this case, $(A\cup B)\setminus B \subsetneq A$. As for your "logical" argument, ...


1

$p \Rightarrow q$ is equivalent to $\neg p \lor q$. So $(p\Rightarrow q)\vee(r \Rightarrow p)\vee(r\Rightarrow s)\vee(r\Rightarrow q)$ is equivalent to $(\neg p \lor q) \lor (\neg r \lor p) \lor (\text{others})$ which is equivalent to $\neg p \lor p \lor (\text{others})$ which is equivalent to $\top$.


3

$$\begin{align} (p\lor q) \to (p \lor r) &\equiv \lnot(p \lor q) \lor p \lor r \\ & \equiv (\lnot p\land \lnot q) \lor p \lor r\\ &\equiv (p \lor \lnot p) \land (p \lor \lnot q) \lor r\\ &\equiv p\lor \lnot q \lor r\\ &\equiv p \lor (q \rightarrow r) \end{align}$$


2

Hint. If $x$ is any variable then $(a\to x)\to(c\to a)$ is false iff $a\to x$ is true and $c\to a$ is false iff $c$ is true and $a$ is false iff $c\to a$ is false. That is, $$(a\to x)\to(c\to a)\quad\Leftrightarrow\quad(c\to a)\ .$$ Use this result twice, and you are finished.


2

$$\begin{align} ((a \rightarrow b) \rightarrow ((a \rightarrow c) \rightarrow (c\rightarrow a))) &\equiv ((\lnot a \lor b)\rightarrow ((\lnot a \lor c) \rightarrow (\lnot c \lor a)))\\ & \equiv \lnot(\lnot a \lor b) \lor (\lnot (\lnot a \lor c)\lor (\lnot c \lor a)\\ &\equiv (a \land \lnot b) \lor (a \land\lnot c) \lor (\lnot c \lor a)\\ & ...


0

You can also use this one: $$(p\lor q)\to (p\lor r) \Longleftrightarrow p\lor(q\to r) $$


1

Hint Use the following equivalence $$A \rightarrow B \Leftrightarrow \neg B \lor A$$ to get a different, equivalent, statement (without the implication symbol).


0

Suppose $\varphi$ is a given formula, and suppose $\Psi$ is any tautology at all. Then, trivially, $(\varphi \land \Psi)$ is equivalent to $\varphi$. So, if you just ask for equivalents to a formal $\varphi$, then I can give you innumerable answers by simply conjoining $\varphi$ with any tautology I like, however complicated. To get an interesting question, ...


2

It depends on what kind of equivalent form you are after. $$\begin{align} \neg((p \wedge q)\to (p \wedge r)) & \equiv \neg ( \neg(p\wedge q)\vee (p\wedge r)) & \text{by Material Implication} \\ & \equiv (p \wedge q) \wedge (\neg p \vee \neg r) & \text{by DeMorgan's Theorem} \\ & \equiv q \wedge (p\wedge (\neg p \vee \neg r)) & ...


2

Proof that WFFs with indistinguishable parentheses have unique decomposition/readability (Credit to GitGud, ProofWiki) Theorem 1 Let A be a WFF of propositional logic. Let S be an initial part of A. Then S is not a WFF of propositional logic. Proof Let l(Q) denote the length of a string Q. By definition, S is an initial part of A iff A=ST for some ...


2

It depends on the formal system you work with. From the looks of it, the formal system you've referred to has no axioms (which some people think problematic). Thus, there doesn't exist any way to pass from Q to (P→Q), without first assuming "P" and then deriving "Q". However, if you have ($\alpha$→($\beta$→$\alpha$)) as an axiom, then from Q as true by ...


0

For iii, "is necessary but not sufficient" can be reworded as "if." Therefore, the answer to iii is Q->P. You can see this if you see that "is necessary but not sufficient" means "the first must be true for the second to be true" which also means "if the second is true than the first was true" which can be simply said "the second implies the first." Part (b) ...


2

"... (Cpp, CpCqCCpCqrCsr} and {CpCqp, CCpCqrCqCpr} were also suffice to show that there is a single axiom basis for a system. ... I'm not sure if his statement applies to any logical calculus, or just implicational calculi." Short answer: True of any logical calculus that has the given theorems. The proof is (as Rezus' proof is) a contructive proof that ...


2

The issue here seems to be not so much propositional calculus as ordinary English. It looks to me as if you're perfectly capable of translating propositional formulas, like $\neg(P\land\neg S)$, by rote substitution (of words and phrases for the connectives and propositional variables) into English sentences like "It is not the case that I will buy the pants ...


3

We know that $\,\lnot P \lor S\,$ essentially defines the implication $P\rightarrow S$: $$\lnot P \lor S \equiv P\rightarrow S$$ This translates to the statement "If I buy the pants, then I'll by the shirt." But we also know that that an implication is equivalent to its contrapositive: $$P\rightarrow S \equiv \lnot S \rightarrow \lnot P$$ This translates ...


2

You can try simplifying the statements as far as possible before translating to English: (a) \begin{align*} (S \lor G) \land (\neg S \lor \neg G) &\equiv (S \land \neg S) \lor (S \land \neg G) \lor (G \land \neg S) \lor (G \land \neg G) \\ &\equiv (S \land \neg G) \lor (G \land \neg S) \\ &\equiv \text{"Either Steve is happy or George is happy ...


2

A = Alice is in the room. B = Bob is in the room. (a) (A ∨ ¬B) ∧ (¬A ∨ B) (b) ¬(A ∧ B) What does your (a) say? Either Alice is in the room and Bob isn't or (vice versa) Alice is isn't in the room and Bob is. In other words, (a) comes to Exactly one of Alice and Bob is in the room. But that isn't what Either Alice or Bob is not in the room ...


2

You might benefit from an answer that explains where you went wrong in the train of thought that led to your answers. Unfortunately, you don't describe your train of thought, and I can't think of a plausible one that would lead to your answers. So I'll just indicate what your answers are saying, without worrying about how you got them. In part (a), your ...


1

(a) "Either Alice or Bob is not in the room" translates to "(Alice is not in the room) OR (Bob is not in the room)". That is: (NOT A) OR (NOT B). (b) "Neither Alice nor Bob is in the room" translates to "(Alice is not in the room) AND (Bob is not in the room)". That is: (NOT A) AND (NOT B).


1

Consider: (a) Alice and Bob are not both in the room. (b) Alice and Bob are both not in the room. I feel that both the statements are equivalent ... Really? Really?? What do you "feel" about (a*) Alice and Bob are not both boys vs (b*) Alice and Bob are both not boys? Or about (a**) Alice and Bob are not both top of the class vs (b**) Alice and Bob ...


1

The statements are not equivalent. For the first statement you can see that not both, this is the opposite as they're both in the room which would be $A\land B$ and negating that gives $\lnot(A\land B)$ For the second statement however, both not means that it has to be true for both of them not being in the room, so it is $(\lnot A) \land(\lnot ...


1

Because there are several partially correct answers above, I thought it would be useful to gather all of the correct answers in one place. Solution 1: The only true statements are D's and B's second statement. In this case the "number of people speaking truth" is either One or Two, depending on whether you count B, who lies once and tells the truth once, ...


0

As the question is How many are Speaking truth?? so, if someone just speaks the truth for once he still will be counted as the one speaking truth My Opinion is that answer is TWO; and those TWO are D and B , yes!both D and B!!! NOTE: Note that world always in the statement of A,..... so, we can easily assume that truth-value of B changes with time, and ...



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