Are the following provable:

a) if, xεΑ===>yεA,then x=y.If not write a counter example

b) (SxT)xU = Sx(TxU) <=====> $\displaystyle S=\emptyset\vee T=\emptyset\vee U=\emptyset$

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- Apr 11th 2009, 05:42 PMxalkset identities
Are the following provable:

a) if, xεΑ===>yεA,then x=y.If not write a counter example

b) (SxT)xU = Sx(TxU) <=====> $\displaystyle S=\emptyset\vee T=\emptyset\vee U=\emptyset$ - Apr 11th 2009, 05:58 PMJhevon
nope, let

$\displaystyle A = \{ x \mid x \in \mathbb{Z}, x \ge 1 \}$

now, consider some integer $\displaystyle x \ge 1$, and take $\displaystyle y = x+ 1$. then we have $\displaystyle x \in A \implies y \in A$, but $\displaystyle x \ne y$

you may also consider the contrapostive. it would probably be more straight forward to you. just make sure you know what the negation of an implication is ;)

Quote:

b) (SxT)xU = Sx(TxU) <=====> $\displaystyle S=\emptyset\vee T=\emptyset\vee U=\emptyset$

- Apr 11th 2009, 07:05 PMxalk
What you proved is:

$\displaystyle (x\in A\Longrightarrow\exists y\in A)\wedge x\neq y$

And not:

$\displaystyle (x\in A\Longrightarrow y\in A)\wedge x\neq y$

The problem is for all x,y.

AS for the 2nd problem ,suppose $\displaystyle S\neq\emptyset\wedge T\neq\emptyset\wedge U\neq\emptyset$,then what?? - Apr 11th 2009, 07:13 PMJhevon
very well. use the contrapositive then. assume x and y are different, does this mean we will have x in A but y not in A?

Quote:

AS for the 2nd problem ,suppose $\displaystyle S\neq\emptyset\wedge T\neq\emptyset\wedge U\neq\emptyset$,then what??

- Apr 12th 2009, 02:34 PMxalk
Here is the proof for the 1st part:

assume xεA===>yεA AND let x=/=y

then put A={x} and so we have:

xε{x} ====> yε{x},but

x=x <===> xε{x},but

x=x and so

yε{x} <===> y=x thus

y=x ,but also x=/=y and we have a contradiction.

Hence x=y

Now for the 2nd proof :

Let [(s,t),u]ε[(SxT)xU] <====> (sεS and tεT)and uεU <=====> sεS and(tεTand uεU) <====> [s,(t,u)]ε[Sx(TxU)].

Now we have to use the 1st proved theorem : (xεA====>yεA) ===> x=y

and :

[(s,t),u] =[s,(t,u)],but also [(s,t),u] =/=[s,(t,u)],hence a contradiction and so:

$\displaystyle S=\emptyset\vee T=\emptyset\vee U=\emptyset$ - Apr 12th 2009, 02:57 PMPlato
Is that really a proof of the original statement?

This is a true statement: $\displaystyle \color{blue}1 \in \emptyset \Rightarrow 2 \in \emptyset$.

But this is a false statement: $\displaystyle \color{red}\left[ {1 \in \emptyset \Rightarrow 2 \in \emptyset } \right] \Rightarrow 1 = 2$.

If you can pick $\displaystyle A=\{x\}$, I can pick $\displaystyle A=\emptyset$.