Question

prove associative law using mathematical method

Answer 1

Assuming this holds, let x=a+(b+c)x=a+(b+c)and y=(a+b)+cy=(a+b)+c. We want to show that x=yx=y, and following the hint we reduce to showing x=xy=yx=xy=y.

I claim that ax=a, bx=b, cx=cax=a, bx=b, cx=c, and likewise for yy. We check for axax:

ax=aa+a(b+c)=a+a(b+c)=aax=aa+a(b+c)=a+a(b+c)=a

Likewise, for bxbx:

bx=ba+b(b+c)=ba+(bb+bc)=ba+(b+bc)=ba+b=bbx=ba+b(b+c)=ba+(bb+bc)=ba+(b+bc)=ba+b=b

The remaining checks are analogous.

Using these identities, you can derive that anything made up of a,b,c,+,.a,b,c,+,. does not change when multiplied by xx, in particular yx=xyx=x:

yx=((a+b)+c)x=(a+b)x+cx=(ax+bx)+cx=(a+b)+c=yyx=((a+b)+c)x=(a+b)x+cx=(ax+bx)+cx=(a+b)+c=y

You can use a symmetric argument to conclude that yx=xy=xyx=xy=x, and hence the claim follows.

For products, you can use a similar trick. Let x=a.(b.c)x=a.(b.c) and y=(a.b).cy=(a.b).c. I claim that x=x+y=yx=x+y=y. To see this, first note that x+a=ax+a=a (because x+a=a+a.(...)=ax+a=a+a.(...)=a. Secondly, x+b=bx+b=b, because

x+b=a.(b.c)+b=a.(b.c)+a.b+a′.b=a.(b.c+b)+a′.b=a.b+a′.b=bx+b=a.(b.c)+b=a.(b.c)+a.b+a′.b=a.(b.c+b)+a′.b=a.b+a′.b=b

(I hope this is legit). Likewise, x+c=cx+c=c. Finally, x+y=yx+y=y, because:

y=(a.b).c=((a+x).(b+x)).(c+x)=(a.b+x).(c+x)=(a.b).c+x=y+xy=(a.b).c=((a+x).(b+x)).(c+x)=(a.b+x).(c+x)=(a.b).c+x=y+x

(I used the identity (u+t).(v+t)=u.v+t.u+t.v+t.t=u.v+t(u+t).(v+t)=u.v+t.u+t.v+t.t=u.v+t). The proof that y+x=xy+x=x is symmetric