Are the following provable:

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

b) (SxT)xU = Sx(TxU) <=====>

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- April 11th 2009, 06:42 PMxalkset identities
Are the following provable:

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

b) (SxT)xU = Sx(TxU) <=====> - April 11th 2009, 06:58 PMJhevon
nope, let

now, consider some integer , and take . then we have , but

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) <=====>

- April 11th 2009, 08:05 PMxalk
- April 11th 2009, 08:13 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 ,then what??

- April 12th 2009, 03:34 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:

- April 12th 2009, 03:57 PMPlato