# Tag: grading policy

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- Stuck trying to simply a logical expression March 5, 2024Simplify: $\neg[p \to \neg(p \land q)] $ My Attempt: 1: $\neg(p \lor \neg(p \land q) \quad \textit{Implication Law} $ 2: $\neg((p \lor \neg p) \lor q)] \quad \textit{First de Morgan's Law} $ 3: $\neg(\neg p \lor \neg q) \quad \textit{Second Idempotent Law} $ I am stuck, and cannot decide on the next step. My mind […]Alix Blaine
- Showing $\Gamma\vdash A$ and $\Gamma\vdash \lnot B$ from $\Gamma\vdash\lnot (A\to B)$ using sequent calculus March 5, 2024Suppose $\Gamma\vdash\lnot (A\to B)$. How do I show that both $\Gamma\vdash A$ and $\Gamma\vdash \lnot B$ using sequent calculus inference rules? Here is my attempt to obtain $\Gamma\vdash A$: From $\Gamma\vdash\lnot (A\to B)$, I can obtain $A\to B,\Gamma\vdash$ using $\lnot$L rule and the cut rule using process outlined in this answer. The sequent $B\vdash B,A\to […]John Davies
- Issues with motivating the conditional connective March 5, 2024The conditional operator $\Rightarrow$ can be tricky to motivate in the cases where it is True. An approach taken by Elliott Mendelson in Number Systems and the Foundations of Analysis is to examine the expression $(C\land D)\Rightarrow D$. My understanding is as follows: Because English sentences that can be paraphrased with this schema are regarded […]MaanDoabeDa
- Is $\emptyset : \emptyset \to \emptyset$ an isomorphism from $(\emptyset, \leq)$ to $(\emptyset, \leq)$? March 4, 2024I was asked to determine whether the following statement is true: If every function $F : P \to P$ is a homomorphism from $(P, \leq)$ to $(P, \leq)$, with $\leq$ an arbitrary order, then $|P| = 1$. It is straightforward to observe that $|P| \not> 1$. However, $|P| = 0$ seems to satisfy the first […]lafinur
- Confused by proof that $L$ is a model of $ZF$ in Jech's set theory March 4, 2024In Chapter 12 of Jech's Set Theory, the model $L$ is defined as: $$ \begin{align} L_0 & = \emptyset, L_{\alpha+1} = \text{def}(L_{\alpha}), \\ L_{\alpha} &= \bigcup_{\beta < \alpha} L_{\beta} \text{ for limit ordinal $\alpha$} \\ L &= \bigcup_{\alpha \in Ord} L_{\alpha} \end{align} $$ where $\text{def}(M) = \{ X \subset M : X \text{ is definable […]Link L
- A sentence $\phi$ with only self-dual connectives corresponds to both truth-functions $f$ and $f^*$ March 4, 2024I'm working on a question for a course in mathematical logic. Prove that if a sentence $\phi$ expresses a truth function $f$ and every connective in $\phi$ is assigned a self-dual truth function, then $\phi$ expresses $f^*$ (the dual of $f$). My lecture notes say that $\phi$ expresses $f$ iff for all $\mathcal{A}:$ $|\phi|_\mathcal{A}=f(|\alpha_1|_\mathcal{A},...,|\alpha_n|_\mathcal{A})$ for […]Amitai
- In the set $P$ of all circles over $\mathbb{R}^2$, ordered by $(P, \subseteq)$, do two circles always have an infimum? March 3, 2024I was aked to determine whether the following statement is true or false. Let $$\mathcal{D} \left( (x_0, y_0), r \right) = \left\{ (x, y) \in \mathbb{R}^2 : (x - x_0)^2 + (y - y_0)^2 \leq r^2 \right\} $$ Let $P = \left\{ \emptyset \right\} \cup \left\{ \mathcal{D}\left( (x_0,y_0), r \right) : x_0, y_0 \in \mathbb{R}, […]lafinur
- Can all undecidable statements on natural numbers be given by Godel's numbering March 3, 2024In the standard proof for Godel incompleteness theorem he code our language with Godel's numbering and of course there are is a lot of freedom in our coding. The proof then provide us with an undecidable statement about the natural numbers. I wonder if it's known whether or not all undecidable statements (maybe up to […]ziv
- Simplifying the logical expression March 3, 2024I stuck on simplifying this expression (¬A)(¬B)(¬C)D + (¬A)(¬B)(¬C)(¬D) + (¬A)(¬B)C(¬D) + (¬A)(¬B)CD + A(¬B)(¬C)(¬D) + A(¬B)C(¬D) + ABC(¬D) Obviously, the first four terms result in (¬A)(¬B) Next I deal with A(¬B)(¬C)(¬D)+A(¬B)C(¬D)+ABC(¬D) = [A(¬B)(¬C)(¬D)+A(¬B)C(¬D)] + [A(¬B)C(¬D)+ABC(¬D)] = A(¬B)(¬D) + AC(¬D) so overall expression is (¬A)(¬B) + A(¬B)(¬D) + AC(¬D) but the solution is (¬A)(¬B) + […]Mikhail Perminov
- What is the motivation behind the definition of the truth table of $p\implies q$? [duplicate] March 3, 2024The truth table of $p\implies q$ is as follows: $$\begin{array}{c|c|c} \boldsymbol{p} & \boldsymbol{q} & \boldsymbol{p\implies q} \\ \hline \color{green}{\text{T}} & \color{green}{\text{T}} & \color{green}{\text{T}} \\ \hline \color{green}{\text{T}} & \color{red}{\text{F}} & \color{red}{\text{F}} \\ \hline \color{red}{\text{F}} & \color{green}{\text{T}} & \color{green}{\text{T}} \\ \hline \color{red}{\text{F}} & \color{red}{\text{F}} & \color{green}{\text{T}} \end{array}$$ I know it's just a definition, but I'd like to […]user1297264

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