You can’t get the lexicographically first proof. You can get a uniformly random one though. (I don’t remember who the argument is due to—probably Vazirani). This is a very rough sketch.
If there are initially N solutions and you add k random constraints, each satisfied by half of all possible proofs, then you expect to have N / 2^k remaining solutions. You can do a binary search on k to find k ~ log N, at which point there are 1-2 solutions. The genie has to say more in the last round, but there is only one thing he can say that satisfies all of the random constraints (you can do that all carefully with a negligible chance of error). Since log N is at most n, the number of bits, the binary search takes log n time.
You can’t get the lexicographically first proof. You can get a uniformly random one though. (I don’t remember who the argument is due to—probably Vazirani). This is a very rough sketch.
If there are initially N solutions and you add k random constraints, each satisfied by half of all possible proofs, then you expect to have N / 2^k remaining solutions. You can do a binary search on k to find k ~ log N, at which point there are 1-2 solutions. The genie has to say more in the last round, but there is only one thing he can say that satisfies all of the random constraints (you can do that all carefully with a negligible chance of error). Since log N is at most n, the number of bits, the binary search takes log n time.