Prove that the states are equivalent if two states are not distinguished by the table filling algorithm.

Prove that the states are equivalent if two states are not distinguished by the table filling algorithm.
Proof: Let us again assume we are talking of the DFA A = (Q, Σ, δ, q₀, F). Suppose the theorem is false; that is, there is at least one pair of states {p, q} such that.
(i) States p and q are distinguishable, in the sense that there is some string w such that exactly one of δ ̂ (p, w) and δ ̂ (q, w) is accepting and yet.
(ii) The table filling algorithm does not find p and q to be distinguished. Call such a pair of states a bad pair.
If there are bad pairs, then there must be some that are distinguished by the shortest strings among all these strings that distinguish bad pair.
Let {p, q} be one such bad pair and let w = a₁, a₂ --- a_n be a string as short as any that distinguishes p from q. Then exactly one of δ ̂ (p, w) and δ ̂ (q, w) is accepting.
Observe first that w cannot be ϵ, since if ϵ distinguishes a pair of states then that pair is marked by the basis part of the table filling algorithm. Thus n ≥ 1.
Consider the states r = δ (p, a₁) and s = δ (q, a₁) states r and s are distinguished by the string a₁, a₂ --- a_n, since this string takes r and s to the states δ ̂ (p, w) and δ ̂ (q, w). However, the string distinguishing r from s is storter than any string that distinguishes a bad pair. Thus, {r, s} cannot be a bad pair.
But the inductive part of the table filling algorithm will not step until it has also inferred that p and q are distinguishable, since it finds that δ (p, a₁) = r and is distinguishable from δ (q,a₁) = s.
We have contradicted our assumption that bad pairs exist. If there are no bad pairs, then every pair of distinguishable states is distinguished by the table filling algorithm and the theorem is true.