Not satisfiable 3SAT instance implications Suppose we have an instance of 3SAT that is NOT satisfiable and we say $S$.
If in $S$ there are the following $8$ clauses
$\left(a\vee b\vee c\right)\wedge\left(a\vee\bar{b}\vee c\right)\wedge\left(a\vee b\vee\bar{c}\right)\wedge\left(a\vee\bar{b}\vee\bar{c}\right)\wedge\left(\bar{a}\vee b\vee c\right)\wedge\left(\bar{a}\vee\bar{b}\vee c\right)\wedge\left(\bar{a}\vee b\vee\bar{c}\right)\wedge\left(\bar{a}\vee\bar{b}\vee\bar{c}\right)$
we have that anyway we set the Boolean value of the variables $a,b,c$, there is a clause that is false.
If in $S$ instead of $a \vee \bar{b} \vee \bar{c}$ we have
$\left(a\vee e\vee f\right)\wedge\left(a\vee\bar{e}\vee f\right)\wedge\left(a\vee e\vee\bar{f}\right)\wedge\left(a\vee\bar{e}\vee\bar{f}\right)$
this implies that anyway we set the Boolean value of the variables $a,b,c,e,f$, there is a clause that is false.
In the first case, we have
$\left(a\vee b\vee c\right)\wedge\left(a\vee b\vee\bar{c}\right)\Rightarrow \left(a\vee b\right)$
$\left(a\vee \bar{b}\vee c\right)\wedge\left(a\vee \bar{b}\vee\bar{c}\right)\Rightarrow \left(a\vee \bar{b}\right)$
$\left(\bar{a}\vee b\vee c\right)\wedge\left(\bar{a}\vee b\vee\bar{c}\right)\Rightarrow \left(\bar{a}\vee b\right)$
$\left(\bar{a}\vee \bar{b}\vee c\right)\wedge\left(\bar{a}\vee \bar{b}\vee\bar{c}\right)\Rightarrow \left(\bar{a}\vee \bar{b}\right)$
and
$\left(a\vee b\right)\wedge\left(a\vee \bar{b}\right)\wedge\left(\bar{a}\vee b\right)\wedge\left(\bar{a}\vee \bar{b}\right)$
is NOT satisfiable.
In the second case, we have
$\left(a\vee e\vee f\right)\wedge\left(a\vee \bar{e}\vee f\right)\Rightarrow \left(a\vee f\right)$
$\left(a\vee e\vee \bar{f}\right)\wedge\left(a\vee \bar{e}\vee\bar{f}\right)\Rightarrow \left(a\vee \bar{f}\right)$
and
$\left(a\vee f\right)\wedge\left(a\vee \bar{f}\right)\Rightarrow a$
and
$a\Rightarrow\left(b\vee c\right)\wedge\left(\bar{b}\vee c\right)\wedge\left(b\vee\bar{c}\right)\wedge\left(\bar{b}\vee\bar{c}\right)$
and this last conjunction is NOT satisfiable.
The question is: given a NOT satisfiable 3SAT instance, if we apply all possible implications and make the conjunction of the right side of the implications, we get a 2SAT instance (the right side of these impliacations is either the conjunction of two variables or a singol variable); is this 2SAT instance not satisfiable? If it isn't, can you give me an example?
You are describing the Resolution refutation system, which is complete in the sense that if a CNF is unsatisfiable, then it can be proven using Resolution. Resolution is also implicitly used in most SAT solvers.
The width of a Resolution refutation is the maximal number of literals in a clause encountered during the refutation. There are 3CNFs on $n$ variables which require width $\Omega(n)$, for example Tseitin contradictions on 3-regular expanders; see for example Massimo Lauria's lecture notes. This means that every Resolution refutation of these 3CNFs must contain a clause having $\Omega(n)$ literals.