Are the following provable:
a) if, xεΑ===>yεA,then x=y.If not write a counter example
b) (SxT)xU = Sx(TxU) <=====> $\displaystyle S=\emptyset\vee T=\emptyset\vee U=\emptyset$
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
this is true, and provable. give a direct proof of the (<=) direction and a proof by contrapositive of the (=>) directionb) (SxT)xU = Sx(TxU) <=====> $\displaystyle S=\emptyset\vee T=\emptyset\vee U=\emptyset$
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??
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?
ok, if they are not empty, then there is some element in each. then we can form the Cartesian products. the elements on the left side are of the form ((s,t),u) for s in S, t in T and u in U, while the right side has elements of the form (s,(t,u)), clearly these are not the sameAS for the 2nd problem ,suppose $\displaystyle S\neq\emptyset\wedge T\neq\emptyset\wedge U\neq\emptyset$,then what??
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:
$\displaystyle S=\emptyset\vee T=\emptyset\vee U=\emptyset$
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 $\displaystyle A=\emptyset$.