Hi everyone,
The review sheet for Exam #2, taking place on Tuesday, 10/25, has been posted under Classroom Resources/Exam Reviews. It is unchanged from the previous year. Let me know if you have any questions.
Best,
Prof. Reitz
Tag: exam 2 review
Logic on Math StackExchange
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- Define something based on the property of another set. April 22, 2024I have always been confused about a logic problem, which I can not form it into a general statement. Here is an example that may express my ideas. We know given a set $A$ and an equivalent relation $\sim$ on it. We can define equivalence class by $\forall x\in A, [x]=\{y,|y \sim x\}$. It seems […]Andrew_Ren
- Prove $\forall x \forall y(xEy \rightarrow \neg x=y)$ in the vocabulary of graphs April 22, 2024Problem: Prove the sentence $\forall x \forall y(xEy \rightarrow \neg x=y)$ in the vocabulary of graphs using the axioms of graph theory. So far: The axioms of graph theory given are antireflexivity and symmetry. Here $xEy$ means there's an edge between the vertices $x$ and $y$. We were told we'd need identity axioms for this […]categoricallystupid
- Proving $\exists x P(x) \rightarrow \forall x P(x)$ from $\forall x\forall y(x=y)$ April 22, 2024Problem: Using identity axioms, prove $\exists x P(x) \rightarrow \forall x P(x)$ from $\forall x\forall y \, x=y$. So far: I'm quite stuck on where to even begin. Working backward, I know we want $P(x)$ and $\exists xP(x)$, but I can't quite see how to write it down. I'm used to proof trees over Fitch-style […]categoricallystupid
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- How is the universal generalization rule used? April 22, 2024The following deduction seems correct: $(x+1)^2=x^2+2x+1$ $\forall x (x+1)^2=x^2+2x+1$ But the following deduction is obviously incorrect: $x+1=2$ $\forall x (x+1)=2$ Did I use the universal generalization rule in a wrong way?William
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- How to construct a type $S$ such that for any $f:S, a:A, b:B$, $f(a,b):T(a)[b/y]$? April 22, 2024Convention: when a type depends on a variable from the context, this variable will be omitted from the type. For example, I will write $\Gamma, x:A \vdash B \text{ type}$ as opposed to $\Gamma, x:A \vdash B(x) \text{ type}$. The notation $B(a)$ will be reserved for the cases when $a$ appears in the premise; e.g. […]user837242
- Enderton Set Theory Question on Cardinals and Ordinals April 21, 2024Assume that A is a set and define (as in Hartogs' theorem) α to be the set of ordinals dominated by A. Show that (i) a is a cardinal number, (ii) card AFushi Wang
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