<p>It's the same. In $λa.(λb.(λc.b))$ brackets are not mandatory because there is no other way to parse $λa.λb.λc.b$ with λ-calculus syntax:</p>

<p>$$t ::= x \mid λx.t \mid t\,t$$</p>

<p>In English, your term is a function which returns a function which returns a function. It's roughly the same as a function which takes three arguments. Except that application must be done in three successive steps. (And in your case, the result is nothing more than the second argument.)</p>

CC BY-SA 3.0
2012-01-27T09:01:34.160Z
2012-01-27T10:10:35.233Z
2012-01-27T10:10:35.233Z
Stéphane Gimenez
<p>(λa.(λb.λc.b))
and
(λa.λb.λc.b)</p>

<p>I was wondering if someone could explain, using mostly English, what that lambda-calculus expression is supposed to mean, and whether there is any difference between the two given above.</p>

CC BY-SA 3.0
2012-01-27T08:32:20.833Z
Welcome to cstheory! This site is about research-level theoretical computer science topics. Please refer to the [FAQ](http://meta.cstheory.stackexchange.com/questions/225/official-faq-for-theoretical-computer-science) to know what kind of questions are suitable for this site.
Mohammad Alaggan
CC BY-SA 3.0
2012-01-27T11:21:44.410Z
The above question would be a better fit at [math.SE]
Artem Kaznatcheev
CC BY-SA 3.0
2012-03-13T05:05:31.443Z
Welcome to cstheory, a Q&A site for research-level questions in theoretical computer science (TCS). Your question does not appear to be a research-level question in TCS. Please see the [FAQ] for more information on what is meant by this and suggestions for sites that might welcome your question. Finally, if your question is closed for being out of scope, and you believe you can edit the question to make it a research-level question, please feel free to do so. Closing is not permanent and questions can be reopened, check the [FAQ] for more information.
Kaveh
<lambda-calculus>
Lambda Calculus - are these two expressions equivalent?
Chuck
# Lambda Calculus - are these two expressions equivalent?

(Asked by: username_0 on 2012-01-27)

(λa.(λb.λc.b))
and
(λa.λb.λc.b)

I was wondering if someone could explain, using mostly English, what that lambda-calculus expression is supposed to mean, and whether there is any difference between the two given above.

**Tags:** lambda-calculus

> **Comments:**
> **username_1**: Welcome to cstheory! This site is about research-level theoretical computer science topics. Please refer to the [FAQ](http://meta.cstheory.stackexchange.com/questions/225/official-faq-for-theoretical-computer-science) to know what kind of questions are suitable for this site.  
> **username_2**: The above question would be a better fit at [math.SE]  
> **username_3**: Welcome to cstheory, a Q&A site for research-level questions in theoretical computer science (TCS). Your question does not appear to be a research-level question in TCS. Please see the [FAQ] for more information on what is meant by this and suggestions for sites that might welcome your question. Finally, if your question is closed for being out of scope, and you believe you can edit the question to make it a research-level question, please feel free to do so. Closing is not permanent and questions can be reopened, check the [FAQ] for more information.  

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## 1 answers

### Answer by username_4 on 2012-01-27. Score: 3

It's the same. In $λa.(λb.(λc.b))$ brackets are not mandatory because there is no other way to parse $λa.λb.λc.b$ with λ-calculus syntax:

$$t ::= x \mid λx.t \mid t\,t$$

In English, your term is a function which returns a function which returns a function. It's roughly the same as a function which takes three arguments. Except that application must be done in three successive steps. (And in your case, the result is nothing more than the second argument.)


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