Consider a classical scalar field theory for a real scalar field $\phi$ given by $$\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)^2-V(\phi)$$ where $V(\phi)$ is the classical potential. In quantum field theory, one defines an effective potential $V_{eff}(\phi)$. And unlike classical field theory where spontaneous symmetry breaking (SSB) is analyzed by minimizing $V(\phi)$, SSB in quantum field theory is analyzed by minimizing $V_{eff}(\phi)$.
For this purpose, one defines a new functional $\Gamma[\phi]$, called the effective action. Intuitively, the name suggests that $\Gamma[\phi]$ must be a modification to the classical action $S[\phi]$ when one takes quantum corrections into account. Indeed when one calculates $\Gamma[\phi]$, one obtains $$\Gamma[\phi]=S[\phi]+\text{quantum corrections of O($\hbar$)}.$$ But that may or may not contain all possible corrections.
However, $\Gamma[\phi]$ is not defined as $$\Gamma[\phi]=S[\phi]+\text{all possible quantum loop corrections}\tag{1}$$ but as $$\Gamma[\phi]=W[J]-\int d^4x j(x)\phi(x).$$
From this definition, how can one be so sure, in general, that evaluation of $\Gamma[\phi]$ gives all possible quantum corrections to $S[\phi]$ in powers of $\hbar$ and nothing is left out? In other words, is there a way to show/see that $(1)$ holds for a generic potential $V(\phi)$?
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