Fairly surprised why no one answered this question..Here's my take on it..

We can either pick numbers or go for reducing the options. I chose reduction..

II is the easiest to evaluate..lets go with that.

a#0 = a+0 - a*0 = a - 0 = a. So II is true. Eliminate all except c and e. So, we need to evaluate only III.

III. (a#b) # c = a # (b#c)

At the first glance this doesn't look right..but need to confirm it!

So we have
[a+b - ab] + c - (a+b-ab)*c = a + [b+c - bc] - a(b+c - bc)
=>a+b - ab +c - [ac + cb -abc] = a +b +c - bc - [ab + ac -abc]
=>a +b -ab +c -ac -cb +abc = a +b +c -bc -ab -ac +abc
=> a+b+c -ab -ac -cb +abc = a +b +c -ab -ac -bc + abc

The two sides _are_ equal! i guess looks can be deceiving! )

E it is! Can someone confirm?

Yes its (E)

vcb wrote:Fairly surprised why no one answered this question..Here's my take on it..

We can either pick numbers or go for reducing the options. I chose reduction..

II is the easiest to evaluate..lets go with that.

a#0 = a+0 - a*0 = a - 0 = a. So II is true. Eliminate all except c and e. So, we need to evaluate only III.

III. (a#b) # c = a # (b#c)

At the first glance this doesn't look right..but need to confirm it!

So we have
[a+b - ab] + c - (a+b-ab)*c = a + [b+c - bc] - a(b+c - bc)
=>a+b - ab +c - [ac + cb -abc] = a +b +c - bc - [ab + ac -abc]
=>a +b -ab +c -ac -cb +abc = a +b +c -bc -ab -ac +abc
=> a+b+c -ab -ac -cb +abc = a +b +c -ab -ac -bc + abc

The two sides _are_ equal! i guess looks can be deceiving! )

E it is! Can someone confirm?