Let R be a relation on a set A Let tR be the transitive clos

Let R be a relation on a set A. Let t(R) be the transitive closure of R. Prove that t(R) = R if and only if R is transitive.

Solution

if RRR then R is transitive

It can be expanded as
if (if (x,y)RR then (x,y)R)then (if (x,y)R and (y,z)R then (x,z)R).

Assume that X is true.
. . . [arguments] . . .
Therefore Y is true.
Let\'s apply this to the above statement. At a very basic level, your proof will be as follows.

Assume that RRR.
. . .
Therefore R is transitive.
to prove that R is transitive The same way, because it is also an if-then statement! So you could develop the previous proof a bit further.

Assume that RRR.
Now suppose that (x,y)R and (y,z)R.
. . .
Therefore (x,z)R.
Hence, R is transitive.
Expanding it we get

Assume that RRR.
That is, if (a,b)RR, then (a,b)R.
In other words, if (a,c)R and (c,b)R for some c, then (a,b)R.
Now suppose that (x,y)R and (y,z)R
Therefore (x,z)R.
Hence, R is transitive.
replacing a,b,c by x,z,y respectively then they are identical. So

Assume that RRR.
That is, if (x,z)RR, then (x,z)R.
In other words, if (x,y)R and (y,z)R for some y, then (x,z)R.
Now suppose that (x,y)R and (y,z)R.
. . .
Therefore (x,z)R.
Hence, R is transitive.
Now again

Assume that RRR,
and suppose that (x,y)R and (y,z)R.
By definition of RR, it follows that (x,z)RR,
but since RRR we have (x,z)R.
We have proved that if (x,y)R and (y,z)R then (x,z)R: hence,R is transitive.