Appendix I: The derivation of \(D_aD_bf\) in Lemma 3
$$\begin{aligned} D_a D_bf= & {} \bigoplus \limits _{ \varsigma =1}^{2}h_{\varsigma }(X_{({n}-2k+1)}^{(n)})\left( \bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{\varsigma }) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i\oplus 1)\right) \\&\oplus \bigoplus \limits _{\begin{array}{c} \varsigma =1 \end{array}}^{2} h_{\varsigma }(X_{({n}-2k+1)}^{(n)}\oplus (a'_{[3]},a'_{[4]}))\left( \bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{\varsigma }) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i\oplus a_i\oplus 1)\right) \\&\oplus \bigoplus \limits _{\begin{array}{c} \varsigma =1 \end{array}}^{2}h_{\varsigma }(X_{({n}-2k+1)}^{(n)}\oplus (b'_{[3]},b'_{[4]}))\left( \bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{\varsigma }) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i\oplus b_i\oplus 1)\right) \\&\oplus \bigoplus \limits _{\begin{array}{c} \varsigma =1 \end{array}}^{2} h_{\varsigma }(X_{({n}-2k+1)}^{(n)}\oplus (a'_{[3]},a'_{[4]})\\&\oplus (b'_{[3]},b'_{[4]}))\left( \bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{\varsigma }) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i\oplus a_i\oplus b_i\oplus 1)\right) \\&\oplus D_{(a_{[1]},a_{[2]})}D_{(b_{[1]},b_{[2]})}g(X^{(n-2k)})\\= & {} \bigoplus \limits _{ \varsigma =1}^{2}h_{\varsigma }(X_{({n}-2k+1)}^{(n)})\left( \bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{\varsigma }) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i\oplus 1)\right) \\&\oplus \bigoplus \limits _{\begin{array}{c} \varsigma =1 \end{array}}^{2} h_{\varsigma }(X_{({n}-2k+1)}^{(n)}\oplus (a'_{[3]},a'_{[4]}))\left( \bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{\varsigma }) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i \oplus 1)\right) \\&\oplus \bigoplus \limits _{\begin{array}{c} \varsigma =1 \end{array}}^{2} h_{\varsigma }(X_{({n}-2k+1)}^{(n)}\oplus (a'_{[3]},a'_{[4]}))\left( D_{a_{[1]}} \bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{\varsigma }) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i\oplus 1)\right) \\&\oplus \bigoplus \limits _{\begin{array}{c} \varsigma =1 \end{array}}^{2} h_{\varsigma }(X_{({n}-2k+1)}^{(n)}\oplus (b'_{[3]},b'_{[4]}))\left( \bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{\varsigma }) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i \oplus 1)\right) \\&\oplus \bigoplus \limits _{\begin{array}{c} \varsigma =1 \end{array}}^{2} h_{\varsigma }(X_{({n}-2k+1)}^{(n)}\oplus (b'_{[3]},b'_{[4]})) \left( D_{b_{[1]}}\bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{\varsigma }) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i\oplus 1)\right) \\&\oplus \bigoplus \limits _{\begin{array}{c} \varsigma =1 \end{array}}^{2} h_{\varsigma }(X_{({n}-2k+1)}^{(n)}\oplus (a'_{[3]},a'_{[4]})\\&\oplus (b'_{[3]},b'_{[4]}))\left( \bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{\varsigma }) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i \oplus 1)\right) \end{aligned}$$
$$\begin{aligned}&\oplus \bigoplus \limits _{\begin{array}{c} \varsigma =1 \end{array}}^{2}h_{\varsigma }(X_{({n}-2k+1)}^{(n)}\oplus (a'_{[3]},a'_{[4]})\\&\oplus (b'_{[3]},b'_{[4]})) \left( D_{a_{[1]}\oplus b_{[1]}}\bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{\varsigma }) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i\oplus 1)\right) \\&\oplus D_{(a_{[1]},a_{[2]})}D_{(b_{[1]},b_{[2]})}g(X^{(n-2k)})\\= & {} \bigoplus \limits _{ \varsigma =1}^{2}D_{(a'_{[3]},a'_{[4]})}D_{(b'_{[3]},b'_{[4]})}h_{\varsigma }(X_{({n}-2k+1)}^{(n)})\left( \bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{\varsigma }) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i\oplus 1)\right) \\&\oplus \bigoplus \limits _{\begin{array}{c} \varsigma =1 \end{array}}^{2} h_{\varsigma }(X_{({n}-2k+1)}^{(n)}\oplus (a'_{[3]},a'_{[4]}))\left( D_{a_{[1]}} \bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{\varsigma }) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i\oplus 1)\right) \\&\oplus \bigoplus \limits _{\begin{array}{c} \varsigma =1 \end{array}}^{2} h_{\varsigma }(X_{({n}-2k+1)}^{(n)}\oplus (b'_{[3]},b'_{[4]})) \left( D_{b_{[1]}}\bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{\varsigma }) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i\oplus 1)\right) \\&\oplus \bigoplus \limits _{\begin{array}{c} \varsigma =1 \end{array}}^{2}h_{\varsigma }(X_{({n}-2k+1)}^{(n)}\oplus (a'_{[3]},a'_{[4]})\\&\oplus (b'_{[3]},b'_{[4]})) \left( D_{a_{[1]}\oplus b_{[1]}}\bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{\varsigma }) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i\oplus 1)\right) \\&\oplus D_{(a_{[1]},a_{[2]})}D_{(b_{[1]},b_{[2]})}g(X^{(n-2k)}). \end{aligned}$$
II. Theorem 4: Proof for the remaining cases when \(|V\cap \Delta |>1\).
(B2) Case \(||V\cap \Delta ||= 2\): In this case there exists one vector \((0_{\frac{n}{2}-k},a_{[2]}, a_{[3]}, 0_{k})\in V\) such that \((a_{[2]}, a_{[3]})\ne 0_{ \frac{n}{2}}\). We have \(||\{(v_1^{(1)},v_4^{(1)}),\ldots , (v_1^{(2^{\frac{n}{2}})}, v_4^{(2^{\frac{n}{2}})})\}||\ge 2^{n/2-1}\) (counting distinct vectors). Assuming, on contrary, that \(||\{(v_1^{(1)},v_4^{(1)}),\dots , (v_1^{(2^{\frac{n}{2}})}, v_4^{(2^{\frac{n}{2}})})\}||< 2^{n/2-1}\) would imply the existence of (at least) 4 vectors
$$\begin{aligned} (v_1^{(i_1)},v_4^{(i_1)}), (v_1^{(i_2)},v_4^{(i_2)}), (v_1^{(i_3)},v_4^{(i_3)}),(v_1^{(i_4)},v_4^{(i_4)}) \in \{(v_1^{(1)},v_4^{(1)}),\dots , (v_1^{(2^{\frac{n}{2}})}, v_4^{(2^{\frac{n}{2}})})\} \end{aligned}$$
such that \((v_1^{(i_1)},v_4^{(i_1)})= (v_1^{(i_2)},v_4^{(i_2)})=(v_1^{(i_3)},v_4^{(i_3)})=(v_1^{(i_4)},v_4^{(i_4)}) \) (since V is a subspace). This contradicts the fact that \(||V\cap \Delta ||= 2\), since the addition of above vectors gives at least 3 vectors of the form \((0_{\frac{n}{2}-k},*,*,0_{k}) \in V \cap \Delta \). Note that
$$\begin{aligned}&\dim (\{v_1^{(1)},\ldots , v_1^{(2^{\frac{n}{2}})} \}) + \dim (\{v_4^{(1)},\ldots , v_4^{(2^{\frac{n}{2}})} \}) \\&\quad \ge \dim (\{(v_1^{(1)},v_4^{(1)}),\ldots , (v_1^{(2^{\frac{n}{2}})}, v_4^{(2^{\frac{n}{2}})}) \}) \ge \frac{n}{2}-1, \end{aligned}$$
so we either have \(\{v_1^{(1)},\dots ,\) \( v_1^{(2^{\frac{n}{2}})}\}= {\mathbb {F}}_2^{n/2-k} \) or alternatively \(\{v_4^{(1)},\dots , v_4^{(2^{\frac{n}{2}})}\}= {\mathbb {F}}_2^{k} \) (up to repetition of some \(v_1^{(i)}\) and \(v_4^{(j)}\)). We consider these two cases separately.
(B2) i) Case \(\{v_1^{(1)},\dots ,\) \( v_1^{(2^{\frac{n}{2}})}\}= {\mathbb {F}}_2^{n/2-k} \): Since \(||\{(v_1^{(1)},v_4^{(1)}),\ldots , (v_1^{(2^{\frac{n}{2}})}, v_4^{(2^{\frac{n}{2}})})\}||\ge 2^{n/2-1}\), we have \(||\{v_4^{(1)},\ldots , v_4^{(2^{\frac{n}{2}})}\}||\ge 2^{k-1}\). There are two subcases to be considered now.
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(1)
When \(\dim (\{(v_3^{(1)},v_4^{(1)}),\ldots , (v_3^{(2^{\frac{n}{2}})}, v_4^{(2^{\frac{n}{2}})}) \})=k\), from item (B1), we have \( D_aD_bf\ne 0\).
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(2)
When \(\dim (\{(v_3^{(1)},v_4^{(1)}),\ldots , (v_3^{(2^{\frac{n}{2}})}, v_4^{(2^{\frac{n}{2}})}) \})=k-1\), we also know \(\dim (\{v_4^{(1)},\ldots ,\) \( v_4^{(2^{\frac{n}{2}})} \})=k-1\) since \(||\{(v_1^{(1)},v_4^{(1)}), \ldots , (v_1^{(2^{\frac{n}{2}})}, v_4^{(2^{\frac{n}{2}})})\}||\ge 2^{n/2-1}\). This can be confirmed using the fact that
$$\begin{aligned} \{(v_1^{(1)},v_3^{(1)},v_4^{(1)}),\ldots , (v_1^{(2^{\frac{n}{2}})},v_3^{(2^{\frac{n}{2}})}, v_4^{(2^{\frac{n}{2}})})\}=2^{n/2-k}\times 2^{k-1} \end{aligned}$$
and because V is a subspace then
$$\begin{aligned} 2^{n/2-k}\times 2^{k-1}\ge & {} \Vert \{(v_1^{(1)},v_3^{(1)},v_4^{(1)}),\ldots , (v_1^{(2^{\frac{n}{2}})},v_3^{(2^{\frac{n}{2}})}, v_4^{(2^{\frac{n}{2}})})\}\Vert \\\ge & {} \Vert \{(v_1^{(1)},v_4^{(1)}),\ldots , (v_1^{(2^{\frac{n}{2}})}, v_4^{(2^{\frac{n}{2}})})\}\Vert \\\ge & {} 2^{n/2-1}. \end{aligned}$$
Further, \( {\mathbb {F}}_2^{n/2-k} \times 0_{2k}\subseteq \{(v_1^{(1)},v_3^{(1)},v_4^{(1)}),\ldots ,\) \( (v_1^{(2^{\frac{n}{2}})}, v_3^{(2^{\frac{n}{2}})},v_4^{(2^{\frac{n}{2}})})\}\), since \(\{v_1^{(1)},\dots ,\) \( v_1^{(2^{\frac{n}{2}})}\}= {\mathbb {F}}_2^{n/2-k} \) and \(\{(v_1^{(1)},v_3^{(1)},v_4^{(1)}),\ldots ,\) \( (v_1^{(2^{\frac{n}{2}})}, v_3^{(2^{\frac{n}{2}})},v_4^{(2^{\frac{n}{2}})})\}\) is a subspace of \({{\mathbb {F}}}_2^{n/2+k}\). Now, since \(\Vert \Phi ^{-1}(H_{1})\Vert \) is odd, there must exist the term \(x_1x_2\cdots x_{\frac{n}{2}-k}h_{1}(X_{({n}-2k+1)}^{(n)})\) in the ANF of f. We also know that there must exist two vectors \(a=(a_{[1]}, a_{[2]}, 0_{2k}),b=(b_{[1]}, b_{[2]}, 0_{2k})\in V\) such that \(a_{[1]}\ne b_{[1]}\), \(a_{[1]}\ne 0_{n/2-k} \) and \(b_{[1]}\ne 0_{n/2-k}\) since \(\{v_1^{(1)},\dots ,\) \( v_1^{(2^{\frac{n}{2}})}\}= {\mathbb {F}}_2^{n/2-k} \). Hence, the term \((h_{1}\oplus h_2)(X_{({n}-2k+1)}^{(n)})\) must be included in the ANF of
$$\begin{aligned} D_{a_{[1]}}D_{b_{[1]}} \bigoplus \limits _{\varsigma =1}^{2}\bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{2}) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i\oplus 1)h_{\varsigma }(X_{({n}-2k+1)}^{(n)}). \end{aligned}$$
Thus,
$$\begin{aligned} \begin{aligned} D_aD_bf=&D_{a_{[1]}}D_{b_{[1]}}\bigoplus \limits _{\varsigma =1}^{2}\bigoplus \limits _{\alpha ^{(\frac{n}{2}-k) }\in \Phi ^{-1}(H_{\varsigma }) }\prod \limits _{i=1}^{\frac{n}{2}-k}(x_i\oplus \alpha _i\oplus 1)h_{\varsigma }(X_{({n}-2k+1)}^{(n)})\\&\oplus D_{(a_{[1]},a_{[2]})}D_{(b_{[1]},b_{[2]})}g(X^{({n}-2k)}) \ne 0\end{aligned}. \end{aligned}$$
(B2) ii) Case \(\{v_4^{(1)},\ldots , v_4^{(2^{\frac{n}{2}})}\}= {\mathbb {F}}_2^{k} \): From item (B1), we directly get \( D_aD_bf\ne 0\).
(B3) Case \(\Vert V\cap \Delta \Vert > 2\): This implies that \(\Vert V\cap \Delta \Vert \ge 4\) and we consider two cases:
$$\begin{aligned} i) \; \dim (\{(v_3^{(1)},v_4^{(1)}), \ldots , (v_3^{(2^{\frac{n}{2}})}, v_4^{(2^{\frac{n}{2}})}) \}) \ge k, \end{aligned}$$
and
$$\begin{aligned} ii) \;\dim (\{(v_3^{(1)},v_4^{(1)}),\ldots , (v_3^{(2^{\frac{n}{2}})}, v_4^{(2^{\frac{n}{2}})}) \})<k. \end{aligned}$$
The case (B3) i) is easily dealt with, since again using (B1) we get \( D_aD_bf\ne 0\).
For the case (B3) ii) we have that \(\dim (\{(v_1^{(1)},v_2^{(1)}),\ldots , (v_1^{(2^{\frac{n}{2}})}, v_2^{(2^{\frac{n}{2}})}) \})> n/2-k\) since
$$\begin{aligned} \dim \{(v_1^{(1)},v_2^{(1)}),\ldots , (v_1^{(2^{\frac{n}{2}})}, v_2^{(2^{\frac{n}{2}})}) \}+\dim \{(v_3^{(1)},v_4^{(1)}),\ldots , (v_3^{(2^{\frac{n}{2}})}, v_4^{(2^{\frac{n}{2}})})\}\ge \dim (V)=\frac{n}{2}. \end{aligned}$$
Further, since V is a subspace and \(\Vert V\Vert =2^{n/2}\), we have
$$\begin{aligned} \Vert S\Vert =\Vert \{(v_1,v_2, 0_{2k}): (v_1,v_2, 0_{2k})\in V \} \Vert >2^{n/2-k}, \end{aligned}$$
for a linear subspace \(S \subset {{\mathbb {F}}}_2^n\) because \(\dim (\{(v_1^{(1)},v_2^{(1)}),\) \(\ldots , (v_1^{(2^{\frac{n}{2}})}, v_2^{(2^{\frac{n}{2}})}) \})> n/2-k\).
Assume now that for any \(v_1,v_2 \) such that \((v_1,v_2, 0_{2k})\in V\), we have that \(v_2\ne 0_{n/2-k} \) implies that \(v_1\ne 0_{n/2-k} \). This would contradict the fact that \(\Vert S\Vert >2^{n/2-k}\). To see this, without loss of generality, we set \(\{ v_1^{i_1},\ldots , v_1^{i_\mu }\}=\{ v_1: (v_1,v_2, 0_{2k})\in S\}\). Since S is a subspace, we must have \(0_{n/2-k}\in \{ v_1^{i_1},\ldots , v_1^{i_\mu }\}\). We also know \( i_\mu \le 2^{n/2-k}<\Vert S\Vert \). Hence, there must exist \((0_{n/2-k}, v_2^{i_\rho },0_{2k})\in V\). Therefore, there must exist one non-zero vector \(a \in V\) of the form \(a=(0_{n/2-k},a_{[2]}, 0_{2k})\in V\). Then, taking some \(b=(b_{[1]},b_{[2]}, 0_{2k})\in V\) we obtain
$$\begin{aligned} \begin{aligned} D_{a}D_bf =&D_{(0_{\frac{n}{2}-k},a_{ [2]})} D_{(b_{[1]},b_{[2]})}g(X^{(n-2k)})\\ =&D_{(b_{[1]},b_{[2]})}(\Phi (X^{(n/2-k)} )\cdot a_{ [2]} )\\ =&D_{b_{[1]}}(\Phi (X^{(n/2-k)} )\cdot a_{ [2]} )\ne 0, \end{aligned} \end{aligned}$$
since \(\Phi \cdot \nu \) has no nonzero linear structure for any \(\nu \in {{\mathbb {F}}}_2^n\setminus \{\mathbf{0 }_{n/2-k}\}\). Combining the cases (A) and (B), we deduce that f does not belong to \({\mathcal {MM}}^{\#}\). \(\square \)