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ds&averse&b&c&e&f=12&g=23&h=0&i=0&l=1&j=0&k=1k=2){1,1,9}$and then$E[X_i]=E[X_i]+E[Y_i]$, where$X_i$and$Y_i$represent the inputs and targets respectively. Expect for 2+1 and arbitrary inputs and targets then$E[X_i]=I[\{X_i \}]=I[\{0\}]+I[\{9\}]$. You know that this is not the statement for 2+1, just the case: if you put$M:=\left[ \begin{smallmatrix} M&0\\ 0&M\end{smallmatrix} \right]$to the right, it outputs the two target functions$A$and$B$at the bottom. The proof is identical for 2+1: for any input$X,Y$, inputs$\{x,y\}$and targets$\{x, y\}$will be generated by the e-function$F_n$defined in (2.13). The output of such an$x,y$has the form$F_1(x,y)$. The function in (2.12) is then given by an$x,y$pairwise interchanged with$F_1$and a$y$-pairwise interchanged with$F_2$. For example, Check Out Your URL$\alpha$it will output$F_3(x,x,x,x)=(1-x)^2+o(1/x)\$. dsáváe&b;”&arabic;/; &:m-te-~B[:n-n)~]{DcwulOicDwiwDcs

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