# Cálculo lambda reducción de funciones

I'm very new to lambda calculus and while I was reading a tutorial , came across with this. Here is my equation.

``````Y = ƛf.( ƛx.f(xx)) ( ƛx.f(xx))
``````

Now if we apply another term, let's say F (YF), then how can we reduce this.If I'm correct according to beta reduction , we can replace all the f in ( ƛx.f(xx)) by ( ƛx.f(xx)), is this correct and if so how can we do that.

Muchas Gracias

preguntado el 04 de julio de 12 a las 03:07

## 1 Respuestas

Reuction steps:

``````Y = ƛf.( ƛx.f(xx)) ( ƛx.f(xx)) = ƛf.( f ( ƛx.f(xx) ƛx.f(xx) ) )
= ƛf.( f ( f (ƛx.f(xx) ƛx.f(xx))))
= ƛf.( f ( f ( f (ƛx.f(xx) ƛx.f(xx))))
= ƛf.( f ( f ( f ( f (ƛx.f(xx) ƛx.f(xx))))) = ...
``````

So this Lambda term goes into an infinite loop...

Explicación:
Let's look on the term `( ƛx.f(xx) ƛx.f(xx) )` nosotros sustituimos `ƛx.f(xx)` con `f'`
lo que significa `(f' f')` => activating the term `f'` en sí mismo.
It might be easier to look at like this:
`( ƛy.f(yy) ƛx.f(xx) )` now when you activate the `ƛy.f(yy)` and provide the input (which substitutes `y` con `ƛx.f(xx)` ) the outcome is: `f(ƛx.f(xx) ƛx.f(xx))` which in turn, can go over the same process again and again and the lambda-expression will only expend...

Observación:
It's wrong to write:
`Y = ƛf.( ƛx.f(xx)) ( ƛx.f(xx))` en realidad debería ser: `Y = ƛf.(ƛx.f(xx) ƛx.f(xx))`
La diferencia entre los `ƛx.f(xx)` y `(ƛx.f(xx))` is that the latter is an activation of `ƛx.f(xx)` - it's meaningless to activate it like this `(ƛx.f(xx))` since we need an `x` (input) to activate it on.

Por último:
`Y = ƛf.( ƛx.f(xx)) ( ƛx.f(xx)) = ƛf.( f ( ƛx.f(xx) ƛx.f(xx) ) )`
`YF = ( ƛx.F(xx)) ( ƛx.F(xx)) = F(ƛx.F(xx)) ( ƛx.F(xx)) = F(YF)`

Respondido 04 Jul 12, 07:07

Could you please tell me the steps to follow or the techniques I may use - Pradeep

I don't get how you got ƛf.( f ( ƛx.f(xx) ƛx.f(xx) ) ) Could you please tell the theorem used - Pradeep

So if we write an activation like YF, is it equals to `( ƛx.F(xx)) ( ƛx.F(xx))` - Pradeep

`( ƛx.F(xx))` is an activation of `ƛx.F(xx)` without parameters (which means nothing on our case cause it can be activated only on a parameter). Not sure what did you mean by YF though. - Nir Alfasi

I came across this while trying to prove that Y is a first point combinator. So my goal is to prove that if F is any term then F(YF) = YF - Pradeep

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