Suppose I am given two transitive closures $R^{+}_1$ and $R^{+}_2$ on the same binary relation $R$ over a set $A$, what steps would I take to show that $R^{+}_1 = R^{+}_2$ ?
$R^{+}_1$ is defined using a basis (if $(x, y) \in R$, then $(x, y) \in R^{+}_1$) and induction (if $(x, y) \in R^{+}_1$ and $(y, z) \in R^{+}_1$, then $(x, z) \in R^{+}_1$). $R^{+}_2$ is defined using a recursive construction. I followed the ProofWiki for a proof for the recursive definition.
Would it suffice to prove each individually, and since they are both closures over the same binary relation $R$, conclude that they are equal?
Let T be the transitive closure of R.
T is the smallest transitive relation containg R.
If T' is a transitive closure of R,
then since closures are the smallest
T subset T' and T' subset T; T = T'.
To show that a relation R for a set S always has a transitive closure, show:
R subset S×S and S×S is transitive;
the intersection of a collection transitive relations is transitive.
With that it is easy to see that
$\cap${ T : T transitive relation for S, R subset T }
is the transitive closure of R.
Afterwards one can show the recursive construction is the closure by showing:
the construction creates a transitive relation containing R;
every transitive relation containing R contains all the ordered pairs of the construction.