Written work – None.
WeBWorK – Assignment #5, due Tuesday, October 10th, at midnight.
OpenLab – OpenLab #5, due Thursday, October 12th, before class.
Logic on Math StackExchange
- Godel's incompleteness theorem: Question about effective axiomatization July 16, 2024I am studying Godel incompleteness theorems and I am struggling with the definition of effective axiomatization. From Wikipedia: A formal system is said to be effectively axiomatized (also called effectively generated) if its set of theorems is recursively enumerable. This means that there is a computer program that, in principle, could enumerate all the theorems […]Tereza Tizkova
- Do we distinguish between these uses of $"="$ in logic? July 16, 2024In the didactics of mathematics it is common (at least in the German literature) to distinguish these uses of "$=$": $(1)$ Identities: $(a+b)^2=a^2+2ab+b^2$ (we sometimes use $\equiv$) $(2)$ Bestimmungsgleichungen (literally: determination equation): $x^2-4x+1=0$ (we sometimes use $\overset{!}=$) $(3)$ Defining equations: $f'(x_0)=\lim\limits_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$ (we sometimes use $:=$) Do we distinguish, on a logical basis, between these […]math_undergrad_questions
- Problem with Fitch : How to Eliminate ¬∃? July 16, 2024fitch problem: I've been working on a Fitch-style proof problem, and I've encountered a difficulty in a specific step (step 11). The problem restricts me to using Taut-Con rules from Con-Rules. Here’s where I’m stuck: I need to eliminate ¬∃ from ¬∃x ¬Cube(x), so I could use Disj.-Elimination, but every attempt I've made results in […]Bogdan Klinowitzki
- Proving that the set of sentences that are true using the symbols $+,<,=$ is the same over all ordered fields July 16, 2024I am interested in whether the set of formulas that one can prove true for a concrete ordered field using the symbols $+, 1 and every x there is y with n · y = x. G ≡ Q. G is o-minimal: all definable subsets of $G$ are finite unions of (possibly degenerate) intervals The […]user1868607
- What strategies can I follow to find the counterexample of an invalid categorical reasoning? [closed] July 15, 2024I am taking the discrete mathematics course in Engineering and I am having problems with the reasoning exercises in the logic part. I have an extremely hard time finding suitable propositional functions and a universal set that invalidates the reasoning, for example with these two invalid reasonings: ∀x: [d(x) ⇒ c(x)]; ∃x: [-c(x) ∧ p(x)] […]Lucio Mazzini
- logic puzzle birth year July 15, 2024Suppose we have two people A and B. A died 129 years after B was born. At least one of A or B was alive for exactly 100 years. B died in 30 B.C. When was A born? This confused me. If B lived 100 years, B would be born in 130 BC and so […]james
- Prove $(B \implies (C \implies D)) \implies (C \implies (B \implies D))$ without the Deduction Theorem July 15, 2024I am reading "Introduction to Mathematical Logic" by Elliott Mendelson, and I am currently at the axiomization of propositional calculus. Mendelson presents the following three axioms (with modus ponens as the only rule of inference): $A1: B \implies (C \implies B)$ $A2: (B \implies (C \implies D)) \implies ((B \implies C) \implies (B \implies D))$ […]gestory2
- If sub-universe $S$ of lattice has congruence $\theta$, does the lattice have a congruence $\lambda = \theta \cap S^2$? [duplicate] July 15, 2024Let $(L, \lor , \land )$ be a lattice and $S$ a sub-universe of the lattice. A sub-universe of a lattice will be any subset of the lattice set that is non-empty and closed under $\land$ and $\lor$. Let $\theta$ a congruence of $(S, \lor_{S\times S}, \land_{S\times S})$. Is it true that there is a […]lafinur
- Is implication true if two statements are always the case? July 14, 2024I have a task that requires me to show that under a certain set of circumstances, a set has property A if and only if it has property B. I can show that under the given circumstances, the set always has property A. I can also show that it always has property B. Since these […]ormondo
- Examples of index set not Turing equivalent to the Halting Problem? July 14, 2024By definition, a set $I \subseteq \mathbb{N} $ is an index set if $\forall i,j ((i \in I \land \varphi_i = \varphi_j) \implies j \in I)$. Thanks to the Rice's Theorem, we know that, said $F$ a family of partial computable functions on the naturals, the set of their code $E = \{e \in \mathbb{N} […]NON
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