The AND function in the lambda is:and = ( a.  b. a b F)I have following expression:and T Tthen 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:( a .  b . a b ( x .  y . y)) ( x .  y . x) ( x .  y . x)( b . a b ( x .  y . y))[a ( x .  y . x)] ( x .  y . x)( b . ( x .  y . x) b ( x .  y . y)) ( x .  y . x)(( x .  y . x) b ( x .  y . y))[b( x .  y . x)]( x .  y . x) ( x .  y . x) ( x .  y . y)( y . x)[x ( x .  y . x)] ( x .  y . y)( y . ( x .  y . x)) ( x .  y . y)( x .  y . x)[y( x .  y . y)]( x .  y . x)TBut when I shortened it like:T T Fthen I've got F, because:T T FT FFWhat am I interpreting wrong?
Why it converts to T
lambda calculus
$T \equiv \lambda x. \lambda y. x$, 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.$$T\ T\ F \equiv (\lambda x. \lambda y. x)\ T\ F \to_\beta(\lambda y. T)\ F \to_\betaT$$ 