Simplification of products of KroneckerDelta[]

Question

What is the reason why the following expression does not simplify to True?

Assuming[
  i > 0 && j > 0 && k > 0 && Element[{i, j, k}, Integers],
  FullSimplify[
    KroneckerDelta[i, j] * KroneckerDelta[i, k] == KroneckerDelta[i, j, k]
  ]
]

(* KroneckerDelta[i, j] KroneckerDelta[i, k] == KroneckerDelta[i, j, k] *)

Is there a corner case where this equation is actually false? Does Mathematica simply not know about this identity? If the latter is the case, how could I define a rule for FullSimplify to simplify products like the one on the left hand side?

Answer 1

You may be able to "teach" KroneckerDelta about this:

Unprotect[KroneckerDelta];
KroneckerDelta /: (Times[KroneckerDelta[i_, j_], KroneckerDelta[k_, m_]] /; i == k) := 
      KroneckerDelta[i, j, m]
Protect[KroneckerDelta];

After the above, here is the new behavior:

KroneckerDelta[a, b] KroneckerDelta[c, d] (* this doesn't match *)
KroneckerDelta[a, b] KroneckerDelta[a, d] (* your use case *)

(* Out:
KroneckerDelta[a, b] KroneckerDelta[c, d]
KroneckerDelta[a, b, d]
*)

Answer 2

Adding PiecewiseExpand[] forces the conversion of KroneckerDelta[] into Piecewise[] equivalents, which play nicer with simplification:

Assuming[i > 0 && j > 0 && k > 0 && {i, j, k} ∈ Integers, 
         KroneckerDelta[i, j] KroneckerDelta[i, k] == KroneckerDelta[i, j, k]
         // PiecewiseExpand // Simplify]

---

Another method consists of converting KroneckerDelta[] expressions into the equivalent Iverson brackets (i.e., Boole[]):

KroneckerDelta[i, j] KroneckerDelta[i, k] == KroneckerDelta[i, j, k] /.
k_KroneckerDelta :> Boole[Equal @@ k] // FullSimplify