We present a deterministic polynomial-time algorithm that solves any instance of 3-SAT, implying $\mathrm{P} = \mathrm{NP}$. The algorithm traverses the implicit binary decision tree $\mathcal{T}_n$ of depth $n$ induced by the Boolean variable space $\{0,1\}^n$. At each of the $n$ levels, instead of searching the subtree exhaustively, it fixes the current variable and verifies whether the resulting partial assignment is consistent with the formula — a check that costs $O(m)$ per level, where $m$ is the number of clauses. The total complexity is $O(n \cdot m)$, which is polynomial in the input size. The key insight is that NP is defined by the existence of a polynomial-time verifier; we show that this verifier, applied to partial assignments at each level of $\mathcal{T}_n$, is sufficient to navigate the tree to a solution without exhaustive search.