Why it converts to T The AND function in the lambda is:

and = (λ a. λ b. a b F)


I have following expression:

and T T


then the result will become T and I am asking me why? 

When I do the beta reduction, then I understand, why I've got T:

(λ a . λ b . a b (λ x . λ y . y)) (λ x . λ y . x) (λ x . λ y . x)
(λ b . a b (λ x . λ y . y))[a →(λ x . λ y . x)] (λ x . λ y . x)
(λ b . (λ x . λ y . x) b (λ x . λ y . y)) (λ x . λ y . x)
((λ x . λ y . x) b (λ x . λ y . y))[b→(λ x . λ y . x)]
(λ x . λ y . x) (λ x . λ y . x) (λ x . λ y . y)
(λ y . x)[x →(λ x . λ y . x)] (λ x . λ y . y)
(λ y . (λ x . λ y . x)) (λ x . λ y . y)
(λ x . λ y . x)[y→(λ x . λ y . y)]
(λ x . λ y . x)
T


But when I shortened it like:

T T F


then I've got F, because:

T T F
T F
F


What am I interpreting wrong?
$T \equiv \lambda x. \lambda y. x$, 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.

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