Assuming `T` is linear:
`y(n)=sum_(k=-oo)^oox(k)T[delta(n-k)]`
If `T` is also time invariant:
`y(n) = T[x(n)] -> y(n-k) = T[x(n-k)]`
If `h(n)` is a response of `T` to `delta(n)` then:
`h(n) = T[delta(n)] -> h(n-k) = T[delta(n-k)]`
`y(n)=sum_(k=-oo)^oox(k)h(n-k)`
But `h(n)` is the impulse reponse of `T`