Timeline for Lambda calculus Beta reduction
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 31, 2016 at 13:53 | answer | added | Josh | timeline score: 1 | |
| Jun 26, 2015 at 13:07 | comment | added | Daniel Fischer | Then you get $\lambda u\, (u\; u)$. Of course instead of $u$, you can call the variable anything you like, e.g. $t$ or $z$. Note that $\lambda z.\, \lambda z\, (z\; z) \equiv \lambda z.\, \lambda y\, (y\; y)$. | |
| Jun 26, 2015 at 13:03 | comment | added | Janitha Tennakoon | it is (λz.λz (z z)) t | |
| S Jun 24, 2015 at 0:24 | history | suggested | iadvd | CC BY-SA 3.0 | added basic mathjax formatting |
| Jun 24, 2015 at 0:00 | review | Suggested edits | |||
| S Jun 24, 2015 at 0:24 | |||||
| Jun 23, 2015 at 13:55 | comment | added | Daniel Fischer | How is λz.λz (z z) t parsed? Is it λz.(λz (z z) t) or (λz.λz (z z)) t? | |
| Jun 23, 2015 at 12:34 | history | edited | Janitha Tennakoon | CC BY-SA 3.0 | deleted 26 characters in body |
| Jun 23, 2015 at 12:19 | review | First posts | |||
| Jun 23, 2015 at 12:20 | |||||
| Jun 23, 2015 at 12:19 | history | asked | Janitha Tennakoon | CC BY-SA 3.0 |