I don't think so. A and B are contradictory. A asserts: (A & !B & !C & !D & !E & !F), B asserts (B & !C & !D & !E & !F) which would imply B and !B therefore A

I don't think so. A and B are contradictory. A asserts: (A & !B & !C & !D & !E & !F), B asserts (B & !C & !D & !E & !F) which would imply B and !B therefore A is false. Only E (or 5 depending on notation) can be true without contradiction.