If program A uses heuristics, program B doesn't need to prove that these heuristics are correct. It only needs to see that A's use of heuristics makes A either output 1 or move on to the brute-force search.

Agreed with the rest of your comment. An easier way to reach the same conclusion: if program B is implemented as "always defect", program A has to run the brute-force search to the end, there's no shortcut (because having any shortcut in the code would make the proofs invalid if B were a copy of A).