Equivalence Relation of $(x\land y)\lor ( \lnot x \land \lnot y)=1$

Question

Define $ R \subseteq T xT $ as follows:

$ (x,y) \in R \iff (x\land y)\lor ( \lnot x\land \lnot y)=1$.

Show, using the laws of Boolean Algebra, that R is an equivalence relation. Hint: if $ A = B = 1 \;then\; A \land B \land C = A \land B \implies C = 1$

I am assuming that I just need to prove that $(x\land y)\lor ( \lnot x\land \lnot y)=1$

So, what I have did so far,

1. $(x\land y)\lor ( \lnot x\land \lnot y) $
2. $(x \lor ( \lnot x\land \lnot y)) \land ( y \lor ( \lnot x\land \lnot y)) $
3. $ (( x \lor \lnot x) \land ( x \lor \lnot y)) \land ((y \lor \lnot x) \land ( y \lor \lnot y)) $
4. $(1 \land (x \lor \lnot y)) \land ((y \lor \lnot x) \land 1)$
5. $(x \lor \lnot y) \land (y \lor \lnot x)$

feel like I am chasing my tail here.

I have go about the question in the wrong direction, I should have prove it to be equivalence relation, which i already know how. My mistake for not writing the full question at first. Thanks

Answer 1

You do not need to prove that $(x\wedge y)\vee(\neg x\wedge\neg y)=1$ (in fact, the latter is false: in general take $x=0$ and $y=1$).

You need to show three things:

1. For all $x$, you have $x~R~x$.
2. For all $x$ and $y$ such that $x~R~y$, you have $y~R~x$.
3. For all $x$, $y$, and $z$ such that $x~R~y$ and $y~R~z$, you have $x~R~z$.

Substituting the definition of $R$, you need to show

1. For all $x$, you have $(x\wedge x)\vee(\neg x\wedge\neg x)=1$.
2. For all $x$ and $y$, if $(x\wedge y)\vee(\neg x\wedge\neg y)=1$, then $(y\wedge x)\vee(\neg y\wedge\neg x)=1$.
3. For all $x$, $y$, and $z$, if $(x\wedge y)\vee(\neg x\wedge\neg y)=1$ and $(y\wedge z)\vee(\neg y\wedge\neg z)=1$, then $(x\wedge z)\vee(\neg x\wedge\neg z)=1$.

The first is just idempotency of boolean operations, the second symmetry. The third requires you to "or" the givens together and apply distributivity. I'll leave the details to you.