Consider this version of MS where we have set $A$ of tasks, $l(a)$, length of each task in $A$ and $m$ number of processors and also a deadline $D$. The question is where we can partition A into m disjoint subsets, $A = A_1 \cup A_2 \cup \ldots \cup A_m$ such that we have: $$ max \left\{ \sum _{a \in A_i} l(a): 1 \le i \le m \right\} \le D ? $$
My attempt(Also hint from Garey & Johnson) consider $m = 2$ and $D = \frac{1}{2} \sum _{a \in A} l(a)$. if so we have $$ max \left\{ \sum _{a \in A_1} l(a), \sum _{a \in A_2} l(a) \right\} \le \frac{1}{2} \sum _{a \in A} l(a) ? $$
this implies that we should have $\sum _{a \in A_1} l(a) = \frac{1}{2} \sum _{a \in A} l(a)$ and also $\sum _{a \in A_1} l(a) = \frac{1}{2} \sum _{a \in A} l(a)$**(if we have such a partition!)**, otherwise we will have a contradiction because if for example $max$ was $\sum _{a \in A_1} l(a)$ we had $$ \sum _{a \in A_1} l(a)< \frac{1}{2} \sum _{a \in A} l(a) \\ \sum _{a \in A_2} l(a)< \frac{1}{2} \sum _{a \in A} l(a) $$ and if we sum the last two inequalities we get this contradiction: $$ \sum _{a \in A} l(a) < \sum _{a \in A} l(a) $$
so now we know that they are equal, and we see that our problem is now an instance of partition(or subset sum with sum equal $B/2$ where $B$ is sum of all elements of A) and we know that partition is $NPC$ so our main problem is $NPC$ too by restriction.
Is my argument correct??
