Is the = sign a higher level than the <-> sign?

I read this in sort-of English this way:

1a) (y=x) is the same as Gy
or
1b) y=(x is the same as Gy)

therefore (I read 'therefore" as "=>", or "implies".)

2) x not-equal y and (z=x) v (z=y)

is the same as

3a) (y=x) is the same as not-Gy)
or
3b) y=(x is the same as not-Gy)

So. From 1a): x=y
From this it follows that 2) is false. (Because 2) says x not-equal y.)
For 2) to be the same as 3), because 2) is false, 3) must be false.
3) contradicts 1). 1) being true, then 3) is false.

Or, a true statement implies that a false statement is the same as a false statement. Which is true.

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2) is nonsensical anyway.

Oh... my bad (I'm portuguese, by the way, so some errors in my english are possible).

<-> is the conective for the logical equivalence. So in the exercise "y = x <-> Gy" is "y is equal to x if and ony if y is G".

3xVy( y = x <-> Gy); therefore 3x3yVz ( ~ (x = y) & (z = x V z = y)) <-> 3xVy( y = x <-> ~ Gy)

=

$\exists x \forall y (y=x \leftrightarrow Gy) \vdash \exists x \exists y \forall z [((\neg(x=y) \wedge (z=x \vee z=y))]\leftrightarrow \exists x \forall y (y=x \leftrightarrow \neg Gy)$

?

Forum has mathjax support.

Yes.

It's precisely that!

Let's try making it a bit more intuitive.

Let the set of all objects be U. We don't know what properties the predicate G has, but we know that, like any unary predicate, it must divide U into a set of objects that satisfy G, call it u1, and a set that doesn't, call it u2.

Then the first statement, call it P1 tells us that only one object satisfies G, ie that u1 consists of exactly one object.

The statement we are asked to prove, call it P2, has two parts, call them A and B, that we are asked to prove are logically equivalent, ie that the <-> holds, given the assumption P1.

Inspecting A carefully, we see that it says that U consists of exactly two objects.
Inspecting B carefully, we see that it says that u2 consists of exactly one object.

Stepping back and thinking of a universe containing only two objects, it seems plausible that P1 would imply that A and B are equivalent. Now we just need to prove it formally.

As is usual with proofs of equivalence (<->), we chould first prove one direction, then the other. Choose the easiest-looking direction first.

So I will show you where currently lies my problem. So, I’m trying a reduction but I’m not seeing how to get the contradiction. I’m guessing that I’ve to get a proposition telling “ ~ (b =a)”, but I’m not seeing how.


https://www.dropbox.com/s/nw0cmgx0bqvybml/IMG_0059.JPG?dl=0