polygraph for the given schedule: (View serializable)
$T_{0}$$\overset{A}{\rightarrow}$ $T_{1}$
$T_{0}$$\overset{A}{\rightarrow}$ $T_{2}$
$T_{0}$$\overset{A}{\rightarrow}$ $T_{3}$
$T_{0}$$\overset{A}{\rightarrow}$ $T_{4}$
$T_{4}$$\overset{B}{\rightarrow}$ $T_{f}$
$T_{1}$ $\rightarrow$$T_{4}$
$T_{2}$ $\rightarrow$$T_{4}$
$T_{3}$ $\rightarrow$$T_{4}$
($T_{1}T_{2}T_{3}$) $\,!$ $\rightarrow$$T_{4}$
$\therefore$ Total 6 cases