SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} ② = F \left(x \left(y ^ {t + 1}, z ^ {t + 1}, v ^ {t}\right), y ^ {t + 1}, z ^ {t + 1}, v ^ {t}\right) - F \left(x \left(y ^ {t + 1}, z ^ {t}, v ^ {t}\right), y ^ {t + 1}, z ^ {t}, v ^ {t}\right) \\ \geq F \left(x \left(y ^ {t + 1}, z ^ {t + 1}, v ^ {t}\right), y ^ {t + 1}, z ^ {t + 1}, v ^ {t}\right) - F \left(x \left(y ^ {t + 1}, z ^ {t + 1}, v ^ {t}\right), y ^ {t + 1}, z ^ {t}, v ^ {t}\right) \\ = \frac {r _ {1}}{2} \langle z ^ {t + 1} + z ^ {t} - 2 x (y ^ {t + 1}, z ^ {t + 1}, v ^ {t}), z ^ {t + 1} - z ^ {t} \rangle . \\ \end{array}
$$
Finally, consider the third part, from the Lemma 3, we get
$$
\circledast = \left\langle \nabla_ {y} F (x (y ^ {t}, z ^ {t}, v ^ {t}), y ^ {t}, z ^ {t}, v ^ {t}), y ^ {t + 1} - y ^ {t} \right\rangle - \frac {L _ {d}}{2} \| y ^ {t + 1} - y ^ {t} \| ^ {2}.
$$
Combining the above inequalities finishes the proof.
Lemma 7 (Proximal descent) For all  $t \geq 0$ , the following inequality holds:
$$
q (z ^ {t}) \geq q (z ^ {t + 1}) + \frac {r _ {1}}{2} \langle z ^ {t} + z ^ {t + 1} - 2 x (z ^ {t}, v (z ^ {t + 1})), z ^ {t} - z ^ {t + 1} \rangle .
$$
Proof From Sion's minimax theorem [54], we have
$$
\begin{array}{l} q (z) = \max  _ {v} p (z, v) = \max  _ {v} \min  _ {x \in \mathcal {X}} \max  _ {y \in \mathcal {Y}} F (x, y, z, v) \\ = \max  _ {v} h (x (z, v), z, v) = \max  _ {v} F (x (y (z, v), z, v), y (z, v), z, v). \\ \end{array}
$$
$$
\begin{array}{l} q \left(z ^ {t}\right) - q \left(z ^ {t + 1}\right) = h \left(x \left(z ^ {t}, v \left(z ^ {t}\right)\right), z ^ {t}, v \left(z ^ {t}\right)\right) - h \left(x \left(z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) \\ \geq h \left(x \left(z ^ {t}, v \left(z ^ {t + 1}\right)\right), z ^ {t}, v \left(z ^ {t + 1}\right)\right) - h \left(x \left(z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) \\ \geq h \left(x \left(z ^ {t}, v \left(z ^ {t + 1}\right)\right), z ^ {t}, v \left(z ^ {t + 1}\right)\right) - h \left(x \left(z ^ {t}, v \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) \\ \geq F (x \left(z ^ {t}, v \left(z ^ {t + 1}\right)\right), y (x \left(z ^ {t}, v \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v \left(z ^ {t + 1}\right)), z ^ {t}, v \left(z ^ {t + 1}\right)) - \\ F (x \left(z ^ {t}, v \left(z ^ {t + 1}\right)\right), y \left(x \left(z ^ {t}, v \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v \left(z ^ {t + 1}\right)) \\ = \frac {r _ {1}}{2} \langle z ^ {t} + z ^ {t + 1} - 2 x (z ^ {t}, v (z ^ {t + 1})), z ^ {t} - z ^ {t + 1} \rangle . \\ \end{array}
$$
E Proof of Proposition 1
From the results in Lemmas 5, 6, and 7, we know that
$$
\begin{array}{l} \Phi (x ^ {t}, y ^ {t}, z ^ {t}, v ^ {t}) \geq \Phi (x ^ {t + 1}, y ^ {t + 1}, z ^ {t + 1}, v ^ {t + 1}) + \left(\frac {1}{c} - \frac {L _ {x} + r _ {1}}{2}\right) \| x ^ {t + 1} - x ^ {t} \| ^ {2} + \\ \left(\frac {r _ {2} - L _ {y}}{2} - L _ {d}\right) \| y ^ {t + 1} - y ^ {t} \| ^ {2} + \frac {(2 - \beta) r _ {1}}{2 \beta} \| z ^ {t + 1} - z ^ {t} \| ^ {2} + \\ \frac {(2 - \mu) r _ {2}}{2 \mu} \| v ^ {t + 1} - v ^ {t} \| ^ {2} + \underbrace {\langle \nabla_ {y} F (x ^ {t + 1} , y ^ {t} , z ^ {t} , v ^ {t}) , y ^ {t + 1} - y ^ {t} \rangle} _ {①} + \\ \underbrace {2 \langle \nabla_ {y} F (x (y ^ {t} , z ^ {t} , v ^ {t}) , y ^ {t} , z ^ {t} , v ^ {t}) - \nabla_ {y} F (x ^ {t + 1} , y ^ {t} , z ^ {t} , v ^ {t}) , y ^ {t + 1} - y ^ {t} \rangle} _ {\text {②}} + \\ \underbrace {2 r _ {1} \langle x (z ^ {t} , v (z ^ {t + 1})) - x (y ^ {t + 1} , z ^ {t + 1} , v ^ {t}) , z ^ {t + 1} - z ^ {t} \rangle} _ {\text {③}}. \\ \end{array}
$$
For the part ①, using the projection update of  $y^{t + 1}$ , we have
$$
\langle \nabla_ {y} F (x ^ {t + 1}, y ^ {t}, z ^ {t}, v ^ {t}), y ^ {t + 1} - y ^ {t} \rangle \geq \frac {1}{\alpha} \| y ^ {t} - y ^ {t + 1} \| ^ {2}.
$$
The part ② is due to the Lipschitz gradient property (see Assumption 1) and error bound (i) in Lemma 4:
$$
\begin{array}{l} 2 \left\langle \nabla_ {y} F \left(x \left(y ^ {t}, z ^ {t}, v ^ {t}\right), y ^ {t}, z ^ {t}, v ^ {t}\right) - \nabla_ {y} F \left(x ^ {t + 1}, y ^ {t}, z ^ {t}, v ^ {t}\right), y ^ {t + 1} - y ^ {t} \right\rangle \\ \geq - 2 L _ {y} \| x \left(y ^ {t}, z ^ {t}, v ^ {t}\right) - x ^ {t + 1} \| \| y ^ {t + 1} - y ^ {t} \| \\ \geq - L _ {y} \sigma_ {6} ^ {2} \| y ^ {t + 1} - y ^ {t} \| ^ {2} - L _ {y} \sigma_ {6} ^ {- 2} \| x (y ^ {t}, z ^ {t}, v ^ {t}) - x ^ {t + 1} \| ^ {2} \\ \geq - L _ {y} \sigma_ {6} ^ {2} \| y ^ {t + 1} - y ^ {t} \| ^ {2} - L _ {y} \| x ^ {t + 1} - x ^ {t} \| ^ {2}. \\ \end{array}
$$
As for the part (3), for any  $\kappa > 0$  it follows that
$$
\begin{array}{l} 2 r _ {1} \left\langle x \left(z ^ {t}, v \left(z ^ {t + 1}\right)\right) - x \left(y ^ {t + 1}, z ^ {t + 1}, v ^ {t}\right), z ^ {t + 1} - z ^ {t} \right\rangle \\ = 2 r _ {1} \langle x (z ^ {t}, v (z ^ {t + 1})) - x (z ^ {t + 1}, v (z ^ {t + 1})), z ^ {t + 1} - z ^ {t} \rangle + 2 r _ {1} \langle x (z ^ {t + 1}, v (z ^ {t + 1})) - x (y ^ {t + 1}, z ^ {t + 1}, v ^ {t}), z ^ {t + 1} - z ^ {t} \rangle \\ \geq - 2 r _ {1} \sigma_ {2} \| z ^ {t + 1} - z ^ {t} \| ^ {2} - \frac {r _ {1}}{\kappa} \| z ^ {t + 1} - z ^ {t} \| ^ {2} - r _ {1} \kappa \| x (z ^ {t + 1}, v (z ^ {t + 1})) - x (y ^ {t + 1}, z ^ {t + 1}, v ^ {t}) \| ^ {2}, \\ \end{array}
$$
where the inequality is from the Cauchy-Schwarz inequality, AM-GM inequality, and error bound (iii) in Lemma 2. Hence,
$$
\begin{array}{l} \Phi \left(x ^ {t}, y ^ {t}, z ^ {t}, v ^ {t}\right) - \Phi \left(x ^ {t + 1}, y ^ {t + 1}, z ^ {t + 1}, v ^ {t + 1}\right) \\ \geq \left(\frac {1}{c} - \frac {L _ {x} + r _ {1}}{2} - L _ {y}\right) \| x ^ {t + 1} - x ^ {t} \| ^ {2} + \left(\frac {1}{\alpha} + \frac {r _ {2} - L _ {y}}{2} - L _ {d} - L _ {y} \sigma_ {6} ^ {2}\right) \| y ^ {t + 1} - y ^ {t} \| ^ {2} + \\ r _ {1} \left(\frac {2 - \beta}{2 \beta} - 2 \sigma_ {2} - \frac {1}{\kappa}\right) \| z ^ {t + 1} - z ^ {t} \| ^ {2} + \frac {(2 - \mu) r _ {2}}{2 \mu} \| v ^ {t + 1} - v ^ {t} \| ^ {2} - \\ r _ {1} \kappa \underbrace {\left\| x \left(z ^ {t + 1} , v \left(z ^ {t + 1}\right)\right) - x \left(y ^ {t + 1} , z ^ {t + 1} , v ^ {t}\right) \right\| ^ {2}} _ {④}. \\ \end{array}
$$
Next, we focus on the negative term. From the fact  $x(z,v) = x(y(z,v),z,v)$  and  $x(y,z,v) = x(y,z,v')$ , the inequality ④ is bounded as follows:
$$
\begin{array}{l} \left\| x \left(z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) - x \left(y ^ {t + 1}, z ^ {t + 1}, v ^ {t}\right) \right\| ^ {2} \\ \leq 2 \| x \left(z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) - x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) \| ^ {2} + 2 \| x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) - x \left(y ^ {t + 1}, z ^ {t + 1}, v ^ {t}\right) \| ^ {2} \\ \leq 2 \| x \left(z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) - x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) \| ^ {2} + 2 \sigma_ {1} ^ {2} \| y ^ {t + 1} - y \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) \| ^ {2}. \tag {14} \\ \end{array}
$$
Here, the second inequality is due to the error bound (i) in Lemma 2. We can further simplify the second error term by (iii)-(iv) in Lemma 4 and (iv), (vii) in Lemma 2:
$$
\begin{array}{l} \left\| y ^ {t + 1} - y \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) \right\| ^ {2} \\ \leq 3 \| y ^ {t + 1} - y \left(z ^ {t}, v ^ {t}\right) \| ^ {2} + 3 \| y \left(z ^ {t}, v ^ {t}\right) - y \left(z ^ {t + 1}, v ^ {t}\right) \| ^ {2} + 3 \| y \left(z ^ {t + 1}, v ^ {t}\right) - y \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) \| ^ {2} \\ \leq 6 L _ {y} ^ {2} \alpha^ {2} \sigma_ {6} ^ {2} \| x ^ {t} - x ^ {t + 1} \| ^ {2} + 6 (\sigma_ {8} + 1) ^ {2} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2} + 3 \sigma_ {3} ^ {2} \| z ^ {t} - z ^ {t + 1} \| ^ {2} + 3 \sigma_ {5} ^ {2} \| v ^ {t} - v _ {+} ^ {t} (z ^ {t + 1}) \| ^ {2}. \tag {15} \\ \end{array}
$$
Plugging (15) into (14), we have
$$
\begin{array}{l} \left\| x \left(z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) - x \left(y ^ {t + 1}, z ^ {t + 1}, v ^ {t}\right) \right\| ^ {2} \\ \leq 2 \| x \left(z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) - x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) \| ^ {2} + 1 2 L _ {y} ^ {2} \alpha^ {2} \sigma_ {1} ^ {2} \sigma_ {6} ^ {2} \| x ^ {t} - x ^ {t + 1} \| ^ {2} + \\ 1 2 \sigma_ {1} ^ {2} (\sigma_ {8} + 1) ^ {2} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2} + 6 \sigma_ {1} ^ {2} \sigma_ {3} ^ {2} \| z ^ {t} - z ^ {t + 1} \| ^ {2} + 6 \sigma_ {1} ^ {2} \sigma_ {5} ^ {2} \| v ^ {t} - v _ {+} ^ {t} (z ^ {t + 1}) \| ^ {2}. \\ \end{array}
$$
Summing the above inequalities and letting  $s_1 \coloneqq \frac{1}{c} - \frac{L_x + r_1}{2} - L_y$ ,  $s_2 \coloneqq \frac{1}{\alpha} + \frac{r_2 - L_y}{2} - L \coloneqq d - L_y \sigma_6^2$  and  $s_3 \coloneqq r_1 \left( \frac{2 - \beta}{2\beta} - 2 \sigma_2 - \frac{1}{\kappa} \right)$ , we have
$$
\begin{array}{l} \Phi \left(x ^ {t}, y ^ {t}, z ^ {t}, v ^ {t}\right) - \Phi \left(x ^ {t + 1}, y ^ {t + 1}, z ^ {t + 1}, v ^ {t + 1}\right) \\ \geq s _ {1} \| x ^ {t + 1} - x ^ {t} \| ^ {2} + s _ {2} \| y ^ {t + 1} - y ^ {t} \| ^ {2} + s _ {3} \| z ^ {t + 1} - z ^ {t} \| ^ {2} + \frac {(2 - \mu) r _ {2}}{2 \mu} \| v ^ {t + 1} - v ^ {t} \| ^ {2} - \\ 1 2 r _ {1} \kappa L _ {y} ^ {2} \alpha^ {2} \sigma_ {1} ^ {2} \sigma_ {6} ^ {2} \| x ^ {t} - x ^ {t + 1} \| ^ {2} - 1 2 r _ {1} \kappa \sigma_ {1} ^ {2} (\sigma_ {8} + 1) ^ {2} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2} - 6 r _ {1} \kappa \sigma_ {1} ^ {2} \sigma_ {3} ^ {2} \| z ^ {t} - z ^ {t + 1} \| ^ {2} - \\ 6 r _ {1} \kappa \sigma_ {1} ^ {2} \sigma_ {5} ^ {2} \| v ^ {t} - v _ {+} ^ {t} (z ^ {t + 1}) \| ^ {2} - 2 r _ {1} \kappa \| x (z ^ {t + 1}, v (z ^ {t + 1})) - x (z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})) \| ^ {2} \\ = \left(s _ {1} - 1 2 r _ {1} \kappa L _ {y} ^ {2} \alpha^ {2} \sigma_ {1} ^ {2} \sigma_ {6} ^ {2}\right) \| x ^ {t} - x ^ {t + 1} \| ^ {2} + s _ {2} \| y ^ {t + 1} - y ^ {t} \| ^ {2} + \left(s _ {3} - 6 r _ {1} \kappa \sigma_ {1} ^ {2} \sigma_ {3} ^ {2}\right) \| z ^ {t + 1} - z ^ {t} \| ^ {2} + \\ \frac {(2 - \mu) r _ {2}}{2 \mu} \| v ^ {t + 1} - v ^ {t} \| ^ {2} - 6 r _ {1} \kappa \sigma_ {1} ^ {2} \sigma_ {5} ^ {2} \| v ^ {t} - v _ {+} ^ {t} (z ^ {t + 1}) \| ^ {2} - 1 2 r _ {1} \kappa \sigma_ {1} ^ {2} (\sigma_ {8} + 1) ^ {2} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2} - \\ 2 r _ {1} \kappa \| x \left(z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) - x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) \| ^ {2}. \tag {16} \\ \end{array}
$$
Moreover, to make the terms in the upper bound consistent, by error bound (iv) in Lemma 4, we have
$$
\begin{array}{l} \left\| y ^ {t + 1} - y ^ {t} \right\| ^ {2} \geq \frac {1}{2} \left\| y ^ {t} - y _ {+} ^ {t} \left(z ^ {t}, v ^ {t}\right) \right\| ^ {2} - \left\| y ^ {t + 1} - y _ {+} ^ {t} \left(z ^ {t}, v ^ {t}\right) \right\| ^ {2} \\ \geq \frac {1}{2} \| y ^ {t} - y _ {+} ^ {t} \left(z ^ {t}, v ^ {t}\right) \| ^ {2} - L _ {y} ^ {2} \alpha^ {2} \sigma_ {6} ^ {2} \| x ^ {t} - x ^ {t + 1} \| ^ {2}. \tag {17} \\ \end{array}
$$
Similarly, we provide a lower bound for  $\| v^{t + 1} - v^t\|^2$  as follows:
$$
\begin{array}{l} \left\| v ^ {t + 1} - v ^ {t} \right\| ^ {2} \geq \frac {1}{2} \left\| v ^ {t} - v _ {+} ^ {t} \left(z ^ {t + 1}\right) \right\| ^ {2} - \left\| v ^ {t + 1} - v _ {+} ^ {t} \left(z ^ {t + 1}\right) \right\| ^ {2} \\ \geq \frac {1}{2} \| v ^ {t} - v _ {+} ^ {t} (z ^ {t + 1}) \| ^ {2} - \mu^ {2} \left(4 L _ {y} ^ {2} \alpha^ {2} \sigma_ {6} ^ {2} \| x ^ {t} - x ^ {t + 1} \| ^ {2} + 4 (\sigma_ {8} + 1) ^ {2} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2} + \right. \\ 2 \sigma_ {3} ^ {2} \| z ^ {t} - z ^ {t + 1} \| ^ {2}). \tag {18} \\ \end{array}
$$
Recalling that  $v_{+}^{t}(z^{t + 1}) = v^{t} + \mu (y(z^{t + 1},v^{t}) - v^{t})$ , the last inequality can be obtained by using error bounds (iv) in Lemma 2 and (iii), (iv) in Lemma 4, i.e.,
$$
\begin{array}{l} \left\| v ^ {t + 1} - v _ {+} ^ {t} \left(z ^ {t + 1}\right) \right\| ^ {2} = \mu^ {2} \left\| y ^ {t + 1} - y \left(z ^ {t + 1}, v ^ {t}\right) \right\| ^ {2} \\ \leq \mu^ {2} \left(2 \| y ^ {t + 1} - y \left(z ^ {t}, v ^ {t}\right) \| ^ {2} + 2 \| y \left(z ^ {t}, v ^ {t}\right) - y \left(z ^ {t + 1}, v ^ {t}\right) \| ^ {2}\right) \\ \leq \mu^ {2} \left(4 L _ {y} ^ {2} \alpha^ {2} \sigma_ {6} ^ {2} \| x ^ {t} - x ^ {t + 1} \| ^ {2} + 4 (\sigma_ {8} + 1) ^ {2} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2} + 2 \sigma_ {3} ^ {2} \| z ^ {t} - z ^ {t + 1} \| ^ {2}\right). \\ \end{array}
$$
Substituting (17) and (18) to (16) yields
$$
\begin{array}{l} \Phi (x ^ {t}, y ^ {t}, z ^ {t}, v ^ {t}) - \Phi (x ^ {t + 1}, y ^ {t + 1}, z ^ {t + 1}, v ^ {t + 1}) \\ \geq \left(s _ {1} - \left(1 2 r _ {1} \kappa \sigma_ {1} ^ {2} + s _ {2} + 2 \mu (2 - \mu) r _ {2}\right) L _ {y} ^ {2} \alpha^ {2} \sigma_ {6} ^ {2}\right) \| x ^ {t + 1} - x ^ {t} \| ^ {2} + \\ \left(\frac {s _ {2}}{2} - (1 2 r _ {1} \kappa \sigma_ {1} ^ {2} + 2 \mu (2 - \mu) r _ {2}) (1 + \sigma_ {8}) ^ {2}\right) \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2} + \\ \left(s _ {3} - (\mu (2 - \mu) r _ {2} + 6 r _ {1} \kappa \sigma_ {1} ^ {2}) \sigma_ {3} ^ {2}\right) \| z ^ {t} - z ^ {t + 1} \| ^ {2} + \left(\frac {(2 - \mu) r _ {2}}{4 \mu} - 6 r _ {1} \kappa \sigma_ {1} ^ {2} \sigma_ {5} ^ {2}\right) \| v ^ {t} - v _ {+} ^ {t} (z ^ {t + 1}) \| ^ {2} - \\ 2 r _ {1} \kappa \| x (z ^ {t + 1}, v (z ^ {t + 1})) - x (z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})) \| ^ {2}. \\ \end{array}
$$
Recalling that  $L_{y} = \lambda L_{x}$ , we have the following results:
- As  $r_1 \geq 2L_x$ , we have  $\sigma_1 = \frac{L_y}{r_1 - L_x} + 1 \leq \frac{L_y}{L_x} + 1 = \lambda + 1$  and  $\sigma_2 = \frac{r_1}{r_1 - L_x} \leq 2$ . As  $r_2 \geq 2L_y$ , similarly, we find  $\sigma_3 = \frac{r_1\sigma_1}{r_2 - L_y} + \frac{\sigma_2}{\sigma_1} \leq \frac{r_1(\lambda + 1)}{L_y} + 2$  and  $\sigma_5 = \frac{r_2}{r_2 - L_y} \leq 2$ . With these
bounds and set  $\kappa := 2\beta$  with  $0 < \beta \leq \min \left\{\frac{24r_1}{360r_1 + 5r_1^2\lambda + (2L_y + 5r_1)^2},\frac{\alpha L_y^2}{384r_1(\lambda + 5)(\lambda + 1)^2}\right\}$  and  $0 < \mu \leq \min \left\{\frac{2(\lambda + 5)}{2(\lambda + 5) + L_y^2},\frac{\alpha L_y^2}{64r_2(\lambda + 5)}\right\}$  we derive that
$$
\begin{array}{l} \frac {(2 - \mu) r _ {2}}{4 \mu} - 6 r _ {1} \kappa \sigma_ {1} ^ {2} \sigma_ {5} ^ {2} \geq \frac {r _ {2}}{2 \mu} - \frac {r _ {2}}{4} - 4 8 (\lambda + 1) ^ {2} r _ {1} \beta \\ \geq \frac {r _ {2}}{\mu} \left(\frac {1}{2} - \frac {\mu}{4} - \frac {L _ {y} ^ {2} \alpha \mu}{8 (\lambda + 5)}\right) \geq \frac {r _ {2}}{4 \mu} \\ \end{array}
$$
$$
\mu (2 - \mu) r _ {2} + 6 r _ {1} \kappa \sigma_ {1} ^ {2} \leq 2 r _ {2} \mu + 1 2 r _ {1} \beta (\lambda + 1) ^ {2} \leq \frac {\alpha L _ {y} ^ {2}}{1 6 (\lambda + 5)}. \tag {19}
$$
With the upper bound (19), we can further bound the coefficient of  $\| z^{t + 1} - z^t\| ^2$  by
$$
\begin{array}{l} s _ {3} - \left(\mu (2 - \mu) r _ {2} + 6 r _ {1} \kappa \sigma_ {1} ^ {2}\right) \sigma_ {3} ^ {2} \\ \geq r _ {1} \left(\frac {1}{\beta} - \frac {1}{2} - 4 - \frac {1}{2 \beta}\right) - \frac {L _ {y} ^ {2}}{1 6 (\lambda + 5)} \left(\frac {r _ {1} ^ {2} (\lambda + 1) ^ {2}}{L _ {y} ^ {2}} + 4 + \frac {4 r _ {1} (\lambda + 1)}{L _ {y}}\right) \\ \geq \frac {r _ {1}}{\beta} \left(\frac {1}{2} - \left(\frac {9}{2} + \frac {r _ {1} (\lambda + 1)}{1 6} + \frac {L _ {y} ^ {2}}{2 0 r _ {1}} + \frac {L _ {y}}{4}\right) \beta\right) \geq \frac {r _ {1}}{5 \beta}. \\ \end{array}
$$
- Let  $\frac{1}{c} - \frac{L_x + r_1}{2} \geq \frac{1}{3c}$  and  $\frac{1}{3c} - L_y \geq \frac{1}{6c}$ . Then, we have  $\frac{1}{c} \geq \max \left\{\frac{3}{4}(L_x + r_1), 6L_y\right\}$ , which implies that  $s_1 \geq \frac{1}{6c}$  and  $\sigma_6 = \frac{2r_1 + \frac{1}{c}}{r_1 - L_x} \geq 2 + \frac{1}{cr_1} \geq \frac{11}{4}$ .
- Let  $\frac{1}{\alpha} - L_y\sigma_6^2 \geq \frac{1}{3\alpha}$  and  $\frac{1}{3\alpha} - L_d \geq \frac{1}{6\alpha}$ . Then, we have  $\frac{1}{\alpha} \geq \max \left\{\frac{3}{2}L_y\sigma_6^2, 6L_d\right\}$  and  $s_2 \geq \frac{1}{6\alpha} + \frac{r_2 - L_y}{2}$ .  $\sigma_8 = \frac{\frac{1}{\alpha} + L_d}{r_2 - L_y} \leq \frac{2}{\alpha r_2} + \lambda + 4$ . With these bounds, if we further assume that  $\frac{1}{\alpha} \geq 5\sqrt{\lambda + 5} L_y$ , we obtain
$$
\begin{array}{l} \frac {s _ {2}}{2} - \left(1 2 r _ {1} \kappa \sigma_ {1} ^ {2} + 2 \mu (2 - \mu) r _ {2}\right) \left(1 + \sigma_ {8}\right) ^ {2} \\ \geq \frac {1}{1 2 \alpha} + \frac {r _ {2} - L _ {y}}{4} - \frac {\alpha L _ {y} ^ {2}}{8 (\lambda + 5)} \left(\frac {4}{\alpha^ {2} r _ {2} ^ {2}} + (\lambda + 5) ^ {2} + \frac {4 (\lambda + 5)}{\alpha r _ {2}}\right) \\ \geq \frac {1}{1 2 \alpha} + \frac {L _ {y}}{4} - \frac {1}{8 (\lambda + 5) \alpha} - \frac {(\lambda + 5) \alpha L _ {y} ^ {2}}{8} - \frac {L _ {y}}{4} \\ \geq \frac {1}{6 0 \alpha} - \frac {(\lambda + 5) \alpha L _ {y} ^ {2}}{8} \geq \frac {1}{9 0 \alpha} \geq \frac {r _ {2}}{1 5} \\ \end{array}
$$
$$
\begin{array}{l} s _ {1} - \left(1 2 r _ {1} \kappa \sigma_ {1} ^ {2} + s _ {2} + 2 \mu (2 - \mu) r _ {2}\right) L _ {y} ^ {2} \alpha^ {2} \sigma_ {6} ^ {2} \geq \frac {1}{6 c} - \frac {\alpha^ {3} L _ {y} ^ {4} \sigma_ {6} ^ {2}}{8 (\lambda + 5)} - \frac {2 L _ {y}}{3} \\ \geq \frac {1}{6 c} - \frac {1 3 L _ {y}}{2 5 0 0} - \frac {2 L _ {y}}{3} \\ \geq \frac {1}{6 c} - \frac {7 L _ {y}}{1 0} \geq \frac {1}{2 4 c} \geq \frac {r _ {1}}{3 2}, \\ \end{array}
$$
where the first inequality is due to  $\frac{2}{3\alpha} \geq L_y \sigma_6^2$  and
$$
s _ {2} \leq \frac {1}{\alpha} + \frac {r _ {2} - L _ {y}}{2} - L _ {y} \sigma_ {6} ^ {2} - L _ {d} = \frac {1}{\alpha} - \frac {r _ {2} + L _ {y}}{2} - L _ {y} \sigma_ {6} ^ {2} - (\sigma_ {1} + 1) L _ {y} \leq \frac {1}{\alpha}.
$$
Putting together all the pieces, we get
$$
\begin{array}{l} \Phi \left(x ^ {t}, y ^ {t}, z ^ {t}, v ^ {t}\right) - \Phi \left(x ^ {t + 1}, y ^ {t + 1}, z ^ {t + 1}, v ^ {t + 1}\right) \\ \geq \frac {r _ {1}}{3 2} \| x ^ {t + 1} - x ^ {t} \| ^ {2} + \frac {r _ {2}}{1 5} \| y ^ {t} - y _ {+} ^ {t} (z ^ {t}, v ^ {t}) \| ^ {2} + \frac {r _ {1}}{5 \beta} \| z ^ {t} - z ^ {t + 1} \| ^ {2} + \frac {r _ {2}}{4 \mu} \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \| ^ {2} - \\ 4 r _ {1} \beta \| x (z ^ {t + 1}, v (z ^ {t + 1})) - x (z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})) \| ^ {2}. \\ \end{array}
$$
F Proof of Proximal Error Bounds and Dual Error Bounds
F.1 Proof of Proposition 2 under Assumption 2
Note that  $h(\cdot, z, v) = \max_{y \in \mathcal{Y}} F(\cdot, y, z, v)$  is  $(r_1 - L_x)$ -strongly convex. Hence, we have
$$
\begin{array}{l} \max _ {v} h (x (z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})), z ^ {t + 1}, v) - h (x (z ^ {t + 1}, v (z ^ {t + 1})), z ^ {t + 1}, v (z ^ {t + 1})) \\ \geq h \left(x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) - h \left(x \left(z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) \tag {20} \\ \geq \frac {r _ {1} - L _ {x}}{2} \| x (z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})) - x (z ^ {t + 1}, v (z ^ {t + 1})) \| ^ {2}. \\ \end{array}
$$
On the other side, by leveraging the KL Assumption 2,  $h(x,z,\cdot)$  satisfies the KL property with same exponent [61, Theorem 5.2]. Thus, we get
$$
\begin{array}{l} \max _ {v} h (x (z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})), z ^ {t + 1}, v) - h (x (z ^ {t + 1}, v (z ^ {t + 1})), z ^ {t + 1}, v (z ^ {t + 1})) \\ \leq \max  _ {v} h \left(x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v\right) - h \left(x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) \tag {21} \\ \leq \frac {1}{\tau} \| \nabla_ {v} h (x (z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})), z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})) \| ^ {\frac {1}{\theta}}, \\ \end{array}
$$
where the first inequality is because
$$
\begin{array}{l} h (x (z ^ {t + 1}, v (z ^ {t + 1})), z ^ {t + 1}, v (z ^ {t + 1})) = \max _ {v} h (x (z ^ {t + 1}, v), z ^ {t + 1}, v) \\ \geq h \left(x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right). \\ \end{array}
$$
Next, we further bound the right-hand part. Recalling the definition of  $v_{+}^{t}(z^{t + 1})$  and with the bound (vii) in Lemma 2, we obtain
$$
\begin{array}{l} \left\| \nabla_ {v} h \left(x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) \right\| \\ = r _ {2} \left\| v _ {+} ^ {t} \left(z ^ {t + 1}\right) - y \left(x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) \right\| \\ = r _ {2} \| \left(1 - \mu\right) v ^ {t} + \mu y \left(z ^ {t + 1}, v ^ {t}\right) - y \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) \| \tag {22} \\ \leq r _ {2} (1 - \mu) \| v ^ {t} - y (z ^ {t + 1}, v ^ {t}) \| + r _ {2} \| y (z ^ {t + 1}, v ^ {t}) - y (z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})) \| \\ \leq r _ {2} \left(\frac {(1 - \mu)}{\mu} + \sigma_ {5}\right) \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \|. \\ \end{array}
$$
Combining (20) (21), and (22), we have
$$
\| x (z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})) - x (z ^ {t + 1}, v (z ^ {t + 1})) \| ^ {2} \leq \frac {2}{(r _ {1} - L _ {x}) \tau} \left(\frac {r _ {2} (1 - \mu)}{\mu} + r _ {2} \sigma_ {5}\right) ^ {\frac {1}{\theta}} \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \| ^ {\frac {1}{\theta}}.
$$
F.2 Proof of Proposition 2 under Assumption 3
By the strong convexity of  $h(\cdot ,z,v) = \max_{y\in \mathcal{Y}}F(\cdot ,y,z,v)$ , we have
$$
\begin{array}{l} h (x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v \left(z ^ {t + 1}\right)) - h \left(x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) \\ \geq h \left(x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) - h \left(x \left(z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) \\ \geq \frac {r _ {1} - L _ {x}}{2} \| x (z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})) - x (z ^ {t + 1}, v (z ^ {t + 1})) \| ^ {2}. \\ \end{array}
$$
On the other side,
$$
\begin{array}{l} h \left(x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) - h \left(x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right), z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) \\ \leq \langle \nabla_ {v} h (x (z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})), z ^ {t + 1}, v _ {+} ^ {t} (z ^ {t + 1})), v (z ^ {t + 1}) - v _ {+} ^ {t} (z ^ {t + 1}) \rangle \\ \leq 2 \operatorname {d i a m} (\mathcal {Y}) r _ {2} \left(\frac {1 - \mu}{\mu} + \sigma_ {5}\right) \| v _ {+} ^ {t} (z ^ {t + 1}) - v ^ {t} \|. \\ \end{array}
$$
Here, the first inequality is because  $h(x,z,\cdot) = \max_{y\in \mathcal{Y}}f(x,y) - \frac{r_2}{2}\| y - \cdot \| ^2$  is concave [52, Theorem 2.26]. The last inequality is from (22) and the fact that  $\| v^{t}\| \leq \mathrm{diam}(\mathcal{V})$ , which could be derived from mathematical induction. Moreover,  $v(z) = \operatorname{argmax}_{v}\left\{\max_{y\in \mathcal{Y}}f(x(y,z,v),y) + \frac{r_1}{2}\| x(y,z,v) - z\|^2 -\frac{r_2}{2}\| y - v\|^2\right\}$  should satisfy  $v(z) = \operatorname{prox}_{\substack{f(x(\cdot ,z,v),\cdot) + \frac{r_1}{2}\| x(\cdot ,z,v) - z\|^2 + 1_{\mathcal{Y}(\cdot)}\\ r_2}}(z(v))$
[52, Theorem 2.26], which indicates that  $\| v(z)\| \leq \mathrm{diam}(\mathcal{V})$ . Thus,
$$
\left\| x \left(z ^ {t + 1}, v _ {+} ^ {t} \left(z ^ {t + 1}\right)\right) - x \left(z ^ {t + 1}, v \left(z ^ {t + 1}\right)\right) \right\| ^ {2} \leq \frac {4 r _ {2} \operatorname {d i a m} (\mathcal {Y})}{r _ {1} - L _ {x}} \left(\frac {1 - \mu}{\mu} + \sigma_ {5}\right) \left\| v _ {+} ^ {t} \left(z ^ {t + 1}\right) - v ^ {t} \right\|.
$$
F.3 Dual Error Bounds
Before presenting the results, we first introduce some useful notation: