SImple NP-hard proof question $3SATplus$

Input: 2 CNF formulas $F_1$, $F_2$ where all clauses have exactly $3$ literals.

Question: Does every truth assignment satisfy at-least as many $F_2$ clauses as $F_1$'s?



Assume $3SATPLUS \in NP$
Prove $3SAT \leq_p$ $3SATplus$
The reduction is as follows
$F_1 = (x \lor \bar{x} \lor x) \land \dots \land (x \lor \bar{x} \lor x)$
For $n$ clauses. This means it will always accept $n$ clauses.
$F_2 = F$
This is polytime because its $2n$ clauses only $n$ more than $F$
if $F \in 3SAT$, then $F_1 = F_2$ since they both satisfy $n$ clauses so $F_1, F_2 \in 3SATplus$
I'll prove the contrapositive
If $F \notin 3SAT$, then $F_1$ satisfies $n$ clauses and $F_2$ can satisfies only up to $n-1$ so $F_1 &gt; F_2$ always. Hence,  $F_1, F_2 \notin 3SATplus$
Is there anything else left to say or does this confirm its $NP-hard$?
If $F\in 3SAT$ it means that there is some assignment that satisfies $n$ clauses. It doesn't mean that every assignment satisfies all clauses. This makes the reduction wrong, since there might be an assignment for $F\in 3SAT$ that satisfies fewer than $n$ clauses. The reduction you've proposed would be valid if you started with $Tautology$, but then, that problem isn't known to be $NP$-complete.

Your problem is $coNP$-complete (as the reduction from $Tautology$ shows), so finding a reduction from $3SAT$ would prove $NP\subseteq coNP$.