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An observed value $k$ must lie in the image of $f$. Choose either preimage $x_k$. The two-to-one function promise and $s\ne 0^n$ imply that its preimages are exactly $x_k$ and $x_k\oplus s$. A computational-basis measurement of the second quantum register projects the joint quantum state onto the unnormalized vectorBy the Born rule, $\mathrm{Pr}(k)=2^{1-n}$. Dividing by its square root gives the conditional first-register pure stateChoosing the other preimage merely interchanges the two summands. The result is a Simon coset state, retaining quantum coherence between the two inputs that share the measured function value. Outcomes outside the image have probability zero and hence no conditional state to normalize.

Apply the Hadamard transform to a Simon coset state. The probability amplitude of $|z⟩$ is $2^{-(n+1)/2}(-1{)}^{x\cdot z}[1+(-1{)}^{s\cdot z}]$. Destructive quantum interference eliminates exactly those $z$ with $z\cdot s=1$. A computational-basis measurement therefore samples the kernel $s^{\perp }$ uniformly, independently of the representative $x$.