- #1

- 493

- 2

Hello!

This is pretty much my first venture in to this notation, and I was wondering if someone could help me with this:

∀x(Fx → ∀y(Fy → y=x))

As far as I am aware, ∀y(Fy → y=x) means for all y, whatever y you plug in, its domain is limited to x, so y is x, and f(y) is f(x): i.e. it limits y to 'being' x.

Something like (Fx & ∀y(Fy → y=x)) makes sense, because its basically saying F can only be over x: there is only one F, namely x; if we tried to introduce something else, ∀y(Fy → y=x) tells us that it has to be x, so it can't be anything else, only x.

If ∀y(Fy → y=x) = u

then what does ∀x(Fx → u) mean? There's no u = x, is it over a domain?, or is it saying if F(x) is true then u? In which case the same notation is saying different things, which apparently it can; but in this case without clarification that this is is happening.

Or is it saying something like: for all x: for F(x), y in F(y) is limited to being x, and that this is true for all y?

Hopefully that makes sense.

Any help appreciated!

This is pretty much my first venture in to this notation, and I was wondering if someone could help me with this:

∀x(Fx → ∀y(Fy → y=x))

As far as I am aware, ∀y(Fy → y=x) means for all y, whatever y you plug in, its domain is limited to x, so y is x, and f(y) is f(x): i.e. it limits y to 'being' x.

Something like (Fx & ∀y(Fy → y=x)) makes sense, because its basically saying F can only be over x: there is only one F, namely x; if we tried to introduce something else, ∀y(Fy → y=x) tells us that it has to be x, so it can't be anything else, only x.

If ∀y(Fy → y=x) = u

then what does ∀x(Fx → u) mean? There's no u = x, is it over a domain?, or is it saying if F(x) is true then u? In which case the same notation is saying different things, which apparently it can; but in this case without clarification that this is is happening.

Or is it saying something like: for all x: for F(x), y in F(y) is limited to being x, and that this is true for all y?

Hopefully that makes sense.

Any help appreciated!

Last edited: