Vtyjkyuk

I have to logically prove

$ b,c = d implies (b=c) ~ wedge (c =d)$

I started with
$$ (forall x)(x in b,c implies x in d ) $$
$$ (b implies d) ~wedge (c implies d) $$

However, I am not sure how to logically change the implication into equality and moreover, how to get the required result.

You have that $b,c=d$ is equivalent to $forall x~(xinb,cleftrightarrow xind)$ by definition of set equality.

You also need premises that $forall x~(xinb,cleftrightarrow (x=bvee x=c))$ and $forall x~(xindleftrightarrow x=d)$ by definition of set construction.

Now generalise to an arbitrary term $a$, raise a few assumptions, use the general biconditionals, and eliminate equalities.

$$deffitch#1#2~~beginarray#1\hline #2endarrayfitchb,c=dforall x~(xinb,cleftrightarrow (x=bvee x=c))\forall x~(xindleftrightarrow x=d)\hdashlineforall x~(xinb,cleftrightarrow xind)\fitch[a]ain b,cleftrightarrow ain d\ain b,cleftrightarrow (a=bvee a=c)\aindleftrightarrow a=d\fitcha=ba=bvee a=c\quadvdots\a=d\b=d\quadvdots\c=d\(b=d)wedge(c=d)\(b=d)wedge(c=d)$$

let $S = b, c$ and $T = d$ and assume $S = T$.

Then we have if $x in S$, then $x in T$. Thus we have $b in T$. Since the only element in T, is $d$, we have $b = d$. This is by the definition of set inclusion. It follows then that $c = d$ so we have $ b = d$ and $c = d$. Thus by transivity of equality, we have $b = c$.