Proposition : The 3-satisfiability problem 3-SAT is NP-complete.
Proof : We give a polynomial-time reduction of SAT to 3-SAT. Let $f(x_1, x_2,\ldots)$ be a Boolean expression. Introduce a variable $y_g$ for each subword $g$ of $f$ that is a Boolean expression (not splitting variables). Now $f$ is satisfiable if and only if the following system is satisfiable:
$y_g = y_{g′} \lor y_{g′′}$ (if $g = (g′ ∨ g′′))$,
$y_g = y_{g′} \land y_{g′′}$ (if $g = (g′ ∧ g′′))$,
$y_g = \neg y_{g′}$ (if $g = \neg g′)$,
$y_f = 1$.
Now $y_g = y_{g′}\lor y_{g′′}$ can be equivalently expressed by: $y_g\lor\neg y_{g′} = 1$, $y_g\lor \neg y_{g′′} =1$, $\neg y_g \lor y_{g′} \lor y_{g′′} = 1$. Similarly, $y_g = y_{g′} \land y_{g′′}$ can be equivalently expressed by: $\neg y_g\lor y_{g′} = 1$, $\neg y_g\lor y_{g′′} = 1$, $y_g\lor\neg y_{g′}\lor\neg y_{g′′} = 1$. The expression $y_g = \neg y_{g′}$ is equivalent to: $y_g \lor y_{g′} = 1,\neg y_g \lor\neg y_{g′} = 1$. By renaming variables, we thus obtain words $w_1,\ldots, w_k$ in $B_2$, such that $f$ is satisfiable if and only if the word $w_1 \land\ldots \land w_k$ is satisfiable.
What does "not splitting variables" mean? Also I don't get the last part
By renaming variables, we thus obtain words $w_1,\ldots, w_k$ in $B_2$, such that $f$ is satisfiable if and only if the word $w_1 \land\ldots \land w_k$ is satisfiable.
It would be useful to see the proof worked on an example to understand what is going on.