It looks like you're new here. If you want to get involved, click one of these buttons!

- All Categories 2.3K
- Chat 499
- Study Groups 18
- Petri Nets 9
- Epidemiology 3
- Leaf Modeling 1
- Review Sections 9
- MIT 2020: Programming with Categories 51
- MIT 2020: Lectures 20
- MIT 2020: Exercises 25
- MIT 2019: Applied Category Theory 339
- MIT 2019: Lectures 79
- MIT 2019: Exercises 149
- MIT 2019: Chat 50
- UCR ACT Seminar 4
- General 67
- Azimuth Code Project 110
- Statistical methods 3
- Drafts 2
- Math Syntax Demos 15
- Wiki - Latest Changes 3
- Strategy 113
- Azimuth Project 1.1K
- - Spam 1
- News and Information 147
- Azimuth Blog 149
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 708

Options

## Comments

\(m+n\). Since \(n=s^n(0)\) and \(m=s^m(0)\), concatenation gives \(s^{n}(s^m(0))=s^{(n+m)}(0)=n+m\). So, concatenation corresponds to summation, and is commutative (since there is only one symbol/function being concatenated).

`\\(m+n\\). Since \\(n=s^n(0)\\) and \\(m=s^m(0)\\), concatenation gives \\(s^{n}(s^m(0))=s^{(n+m)}(0)=n+m\\). So, concatenation corresponds to summation, and is commutative (since there is only one symbol/function being concatenated).`