For the below VHDL code snippet not relating to the previous state machine problems, what of the choices below would be a SUPER critical addition to make this work, without which the intended state diagram transitions would never likely occur.

case Q is

when $"1101"=>$ if $($T $=\text{'}1\text{'}$ F $=\text{'}1\text{'})$ then Q$<="0101"$;elsif

$($T $=\text{'}0\text{'}$ and R $=\text{'}0\text{'}$ and F $=\text{'}0\text{'})$ then Q$<="1010"$;end if;

when $"0101"=>$ if $($T $=\text{'}1\text{'}$ and S $=\text{'}1\text{'})$ then Q$<="1101"$;elsif

$($S $=\text{'}0\text{'}$ and R$=\text{'}1\text{'})$ then Q$<="0101"$;elsif

$($S $=\text{'}0\text{'}$ and R $=\text{'}0\text{'})$ then Q$<="1010"$;end if;

when $"1010"=>$ if $($T $=\text{'}1\text{'}$ and R $=\text{'}0\text{'})$ then Q$<="0101"$;elsif

$($T $=\text{'}0\text{'}$ or R $=\text{'}1\text{'})$ then Q$<="1111"$;end if;

when $"1111"=>$ if $($S $=\text{'}1\text{'}$ and T$=\text{'}0\text{'})$ then Q$<="1010"$;elsif

$($S $=\text{'}0\text{'}$ and R $=\text{'}1\text{'})$ then Q$<="1101"$;end if;

end case;