Why it converts to T
$T \equiv \lambda x. \lambda y. x$, which means it is a $\lambda$-function of two arguments, returning the first one, so $(T\ T\ F)$ $\beta$-reduces to $T$, which can be shown more formally as follows.

$$T\ T\ F \equiv (\lambda x. \lambda y. x)\ T\ F \to_\beta
(\lambda y. T)\ F \to_\beta
T$$