# Why Cook-Levin thorem's proof can mean SAT's NP-Hardness

I'm studying about Cook-Levin theorem but there is a problem I faced. Cook-Levin theorem shows that any NPTM can be encoded as a boolean formula. About given language $$A$$, instance $$w$$, and NPTM $$M$$ that decides $$A$$, I understand that when a boolean formula $$φ$$ is true when $$M$$ accepts $$w$$ and false when $$M$$ rejects $$w$$, decision problem $$w∈A?$$ will be the same problem as $$φ=true?$$. But I am confused that can we actually certificate existence of reduction from any other NP problem to SAT, with Cook-Levin theorem. I can't make actual Karp reduction to SAT that doesn't picky about where the reduction comes from, so I can't tell that SAT is in NP-Hard. Please help me to understand why conversion to boolean formula can mean SAT's NP-Hardness...

Any language $$L$$ in NP is decided by some nondeterministic Turing machine $$M$$. By Cook–Levin, the problem "Does $$M$$ accept input $$x$$?" can be decided by constructing a Boolean formula $$\varphi_{M,x}$$ that is satisfiable if, and only if, $$M$$ accepts $$x$$. Hence $$L$$ reduces to SAT, so SAT is NP-hard.