N. Dunfield, P. Of-lemma, and B. , We define some additional elements of ? by g = aBABB h = abbAb n = aBBAB m = aBaab v = ABAAb By symmetry, we can assume g ? P . We now try all the possibilities for whether the elements {g, h, n, d, c, m, v} are in P or not, in each case leading to the contradiction that 1 ? P . Case h ? P : Case n ? P : Case d ? P : Case c ? P : Case m ? P : Then P contains the following, which is 1 in ?: cgndhmgmcmdhmchm Case M ? P : Then P contains the following, which is 1 in ?: MgndhdMgndhdMgnMhdMndMdMgndhdMgnMhdMgn Case C ? P : Then P contains the following, which is 1 in ?: hCggnhCgChCgd Case D ? P : Case c ? P : Case m ? P : Then P contains the following, which is 1 in ?: mcDnDmcDnDmcmnDmhDmDmcDnDmcmnDmc Case M ? P : Then P contains the following, which is 1 in ?: gnMDnMgnnMgncDnMg Case C ? P : Then P contains the following, which is 1 in ?: ChDnhCgDnhCggnnhCggnhCg Case N ? P : Case d ? P : Then P contains the following, which is 1 in ?: NdhNgdNdhhNgdNdhdNdhNgdNgdNdhdNdhNgdN Case D ? P : Case c ? P : Case m ? P : Then P contains the following, which is 1 in ?: DmchNgmgmDNm Case M ? P : Case f ? P : Then P contains the following, which is 1 in ?: hNcNfhMhNcNfhNhNfDf Case F ? P : Then P contains the following, which is 1 in ?: NhNcFMhNhNcFMhNcMhNcFcF Case C ? P : Case v ? P : Case f ? P : Then P contains the following, which is 1 in ?: ChCgChNhCgChVChChNhCgChhChNhCgChVChVhhChNhCgChVChV Case H ? P : Case n ? P : Case d ? P : Then P contains the following, which is 1 in ?: nHggdHnHgnnHggnnHggdHnHgndgnnHggnnHggdHnHgnn

D. Case and ?. Case, which is 1 in ?: DnHcHDHgnnHcHDHDnnHcHDHDnDH Case C ? P : Then P contains the following, which is 1 in ?: DHCnHgnnHggnnHgnHCnHgnnCnHCnHgnn Case N ? P : Case d ? P : Case c ? P : Then P contains the following, which is 1 in ?: NNgdHcNgdHggdHgdNNgdHcNgdHggNgdHcNgdNNgdHcd Case C ? P : Then P contains the following, which is 1 in ?: NgdHggdCgdNNgdCgNgdHggdCgdHggdCgdNd Case D ? P : Case c ? P : Case m ? P : Then P contains the following, which is 1 in ?: NcHcHDNmDHDcHcHDNmHcHDNcHccHD Case M ? P : Case v ? P : Then P contains the following, which is 1 in ?: HcHDMvHcHDNcHHcHDNcHcv Case V ? P : Then P contains the following, Then P contains the following which is 1 in ?: DcVcHDNcHcHDcVcHDHVcHDH Case C ? P : Case m ? P : Then P contains the following, which is 1 in ?: DmgmDNmDNgmgmDNmDHDDmgmDNmDHHgmDNm Case M ? P : Then P contains the following, which is 1 in ?: HDHCMHCMHDMCDHCMHDMHDMCMHCMHDM References

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