@jpwel.bsky.social
o4-mini-high is thinking along the right lines, but still makes a mistake. It says "using Rule 2 (k) times and Rule 3 (m) times (in any order), the total B‑count will be 2^k - 3m".
This is true only if the k rule 2s all precede the m rule 3s - not "in any order". A correct proof is more subtle.
This is true only if the k rule 2s all precede the m rule 3s - not "in any order". A correct proof is more subtle.
April 30, 2025 at 2:44 PM
o4-mini-high is thinking along the right lines, but still makes a mistake. It says "using Rule 2 (k) times and Rule 3 (m) times (in any order), the total B‑count will be 2^k - 3m".
This is true only if the k rule 2s all precede the m rule 3s - not "in any order". A correct proof is more subtle.
This is true only if the k rule 2s all precede the m rule 3s - not "in any order". A correct proof is more subtle.