$A\leq_p B$ means that $A$ is polynomial time reducible to $B$. ($B$ is solvable implies $A$ is solvable.)

Note that by the contrapositive, if we suspect that $A$ cannot be solved in poly time, then $B$ is most likely not solvable either. ($A$ is not solvable implies $B$ is not solvable).

$A$ is a problem in $NP$ if for all $\Phi\in A$, $\exists w, |w|=\mathcal{O}(n^{\mathcal{O}(1)})$ and there exists a poly time Boolean relative to $w$ such that $x\in L$ if $R(\Phi,w)=1$.

Theorem: 3 SAT is NP complete:

Must show that 3SAT is NP and NP-Hard. Also, we could show that circuit sat is poly time reducible to 3SAT.