Proof: Induction with respect to $n$. First, let us denote $A_i^{(m)} = (a_{i1}, a_{i2},\dots,a_{im}),\, m \le n $. Then $a_{i+1,j}$ depends only on $A_i^{(j)}$. Suppose that the assertion is true for $n \le N$. Let $n=N+1$. Because $a_{i+1,j}$ depends only on $A_i^{(j)}, \, j=1,2,\dots,N $, then there exist $k$ such that $A_{k+2}^{(N)}= A_{k}^{(N)}$. Then $A_{k+3}^{(N)}= A_{k+1}^{(N)}$. Denote $A^{(N)}=A_{k}^{(N)}, \, B^{(N)}=A_{k+1}^{(N)}$. It holds: $A^{(N)}=A_{k+2l}^{(N)}, \, B^{(N)}=A_{k+2l+1}^{(N)}, \, l=0,1,\dots$ Let $ f(A_s^{(N)}, a_{s,N+1})= Card \{a_{s,l}|a_{s,l} \ge a_{s,N+1},\, 1\le l \le N\}$. Notice that if $X^{(N)}$ is fixed, $f(X^{(N)}, x)$ is decreasing as a function of $x$. Let $x=a_{k,N+1}$. $ a_{k+2, N+1}(x)=\phi(A_k^{(N)}, x) = f(B_k^{(N)}, f(A_k^{(N)}, x) ) $. It is easy to see that $\phi$ is increasing as a function of $x$, because it is a superposition of two decreasing functions. Let assume that $x \ge \phi(x)$. It follows $ x \ge \phi(x) \ge \phi^2(x) \ge \dots $ Because $\phi$ is bounded it will stabilize at the end. The case $x < \phi(x)$ is similar. Thus, there exists $m$, so that $ \phi^m(x) = \phi^{m+1}(x) \Rightarrow a_{k+2m, N+1}=a_{k+2m+2, N+1}$ or for $ k_1= k+2m,\,\Rightarrow \, A_{k_1+2}= A_{k_1}$.