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Using the summation convention and the Kronecker delta ($\delta$), one can show that $$a_i\delta_{ij} = a_j.$$

If one expands the expression, one is looking at $$ a_i\delta_{ij} = a_1\delta_{1j} + a_2\delta_{2j} + \cdots + a_n\delta_{nj}. $$

In the above $i$ is usually called a 'dummy' variable and 'j' a free variable. The slightly confusing thing that happens is that the free variable seems to 'jump' from the $\delta$ to the $a$ although it does not originally figure as a subscript of it. One does not (at least as a beginner) see this coming.

To 'prove' this, one could observe the pattern

\begin{align} j=1 \implies& a_i\delta_{i1} = a_1 \implies a_i\delta_{ij} = a_j & \text{(since j=1)} \\ j=2 \implies& a_i\delta_{i2} = a_2 \implies a_i\delta_{ij} = a_j & \text{(since j=2)} \\ & \cdots \\ j=n \implies& a_i\delta_{in} = a_n \implies a_i\delta_{ij} = a_j & \text{(since j=n)} \end{align} So one can conclude that this holds for arbitrary $j$. But what is this kind of proof? It is not a proof by induction (or is it?), at the same time, one has the $\cdots$ middle part which seems like hand waving. One could talk about 'arbitrary j', but what comes close to such a proof formally?

If one were a purist, and introduced this notation in an overly rigorous book on set theory, what would the proof look like Hilbert style? Alternatively, what would the proof look like in a proof assistant like CoQ? What 'theorems', 'definitions', would be it using?

Maybe this is not a proof but a 'verification'? If it was, how would an overly rigorous go about proving this as an identity?

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  • $\begingroup$ In your "proof," you fixed $j$ at specific values and deduced correctly that $a_i\delta_{ij}=a_j$. Was there anything special about $j=1$, $j=2$, or $j=n$? Simply fix $j$ at any $j_0$, $1\le j_0\le n $, and deduce that $a_i\delta_{ij_0}=a_{j_0}$. $\endgroup$ – Mark Viola Aug 20 '17 at 15:04
  • $\begingroup$ How is this 'fixing' expressed in let's say a formal Hilbert-system proof or in a formal natural deduction proof? Same question for the '...' $\endgroup$ – weakmoons Aug 20 '17 at 15:09
  • $\begingroup$ Actually, I do not understand your problem. You are summing over $i$, and all summands are zero (because of the definition of $\delta_{ij}$) with the exception of the summand where $i=j$. So the result is clearly $a_j \delta_{jj}=a_j$. $\endgroup$ – Francesco Polizzi Aug 20 '17 at 15:12
  • $\begingroup$ My problem is that this does not amount to a proof in the formal sense as far as I can see. By formal sense I mean, as an example, the proof tree of P & Q in: math.stackexchange.com/questions/374937/…. If this is a proof, how does its proof tree look like? $\endgroup$ – weakmoons Aug 20 '17 at 15:17

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