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Proof step 4 from proof abs-dirs-deldatakeys-1 (previous step)

⊦ (inodekey(ino) ∈ ri \ datakeys(key, ri) ∧ fs[(∅)[inodekey(ino)]].directory → inodekey(ino) ∈ ri) ∧ ( inodekey(ino) ∈ ri ∧ fs[ri[inodekey(ino)]].directory → mkdir (fs[ri[inodekey(ino)]].size, fs[ri[inodekey(ino)]].meta, fs[ri[inodekey(ino)]].nlink, fs[ri[inodekey(ino)]].nsubdirs, abs-children(ino, fs, ri)) = mkdir (fs[ri[inodekey(ino)]].size, fs[ri[inodekey(ino)]].meta, fs[ri[inodekey(ino)]].nlink, fs[ri[inodekey(ino)]].nsubdirs, abs-children(ino, fs, ri \ datakeys(key, ri))))

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: inodekey(ino) ∈ ri

Additional information

Used simplifier rules: D-inj: ⊦ mkdir(n0, md0, m0, m1, entries0) = mkdir(n1, md1, m2, m3, entries1) ↔ ¬(md0 ≠ md1 ∨ n0 ≠ n1 ∨ m0 ≠ m2 ∨ m1 ≠ m3 ∨ entries0 ≠ entries1);%C-ext: ⊦ entries = abs-children(ino, fs, ri) ↔ (∀ str. (str ∈ entries ↔ dentrykey(ino, str) ∈ ri) ∧ ( str ∈ entries → entries[str] = fs[ri[dentrykey(ino, str)]].ino));%in-children: ⊦ str ∈ abs-children(ino, fs, ri) ↔ dentrykey(ino, str) ∈ ri;%at-children: ⊦ dentrykey(ino, str) ∈ ri → abs-children(ino, fs, ri)[str] = fs[ri[dentrykey(ino, str)]].ino;% ⊦ key ∈ ri \ ks ↔ key ∈ ri ∧ ¬ key ∈ ks; (ws)%⊦ key ∈ datakeys(key0, ri) ↔ key ∈ ri ∧ key.data? ∧ key0.inode? ∧ key.ino = key0.ino; (s)% ⊦ ¬ dentrykey(m, str).data?; (s)%⊦ (ri \ datakeys(key, ri))[dentrykey(ino, str)] = ri[dentrykey(ino, str)]; (s) Used formulas: inodekey(ino) ∈ ri