P!=NP

Let `B(N)` be the set of boolean strings with length N. Example:

    B(3) = [[000], [001], [010], [011], [100], [101], [110], [111]]

Let `F` be a set of boolean functions of arity N that tests if `a0, a1, ... aN` is a specific member of B(N).  Example:

    F(3) = [f0,f1,f2,f3...] where
        f0(a,b,c) = !a & !b & !c,
        f1(a,b,c) = !a & !b & c,
        f2(a,b,c) = !a & b & !c,
        f3(a,b,c) = !a & b & c,
        ... etc...

Let `f` be a boolean function of arity N that tests its inputs against each `g` in `F`. For example, if N=3, a value of `f` could be:

    f(a,b,c) = f0(a,b,c) & f1(a,b,c) & !f2(a,b,c) ...

In order to find `a, b, c` such that `f(a,b,c) = 1`, then you must find `a,b,c` such that `f0(a,b,c) = 1`, `f1(a,b,c) = 2`, `f2(a,b,c) = 3`, etc. Considering there are `2^N` elements in `B(N)`, then, you must, at least, execute `2^N` tests. Thus, SAT can't be solved in Polynomial time, and P!=NP.

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