# Polynomial Reduction from $3SAT$ to $MSAT$

I am supposed to show that

$$3SAT$$ $$\rightarrow$$ Every clause hast exact $$3$$ literals

is polynomial reducible to

$$MSAT$$ $$\rightarrow$$ At least half of every clauses' literals are true

Let $$F$$ be a fulfilling configuration of $$3SAT$$, i.e. $$F$$ = $$C_1 \wedge ... \wedge C_n$$, where $$C$$ contains exactly $$3$$ literals. I tried to reduce $$3SAT$$ to $$MSAT$$ by introducing $$2$$ new variables $$a_1$$ and $$a_2$$ and add them to each clause of $$F$$. This reduction formula would then be $$f(F) = g(C_1) \wedge ... \wedge g(C_n)$$ with $$g(C) = l_1 \vee l_2 \vee l_3 \vee a_1 \vee a_2$$. Now I am a little bit stuck, could I simply consider that $$a_1$$ and $$a_2$$ are always true? Or do I have distinguish between the $$4$$ cases where $$a_1$$ and $$a_2$$ are true, $$a_1$$ is not true, but $$a_2$$ is true and so on? If so, it is not possible that half of the literals of each clause are true anymore. In this case it would be not a valid $$MSAT$$ instance?