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Proof step 99 from proof afs-cons-rename-file-negdentry (previous step)

(ino = m0 ⊃ mkdir(m1 + 1, md1, # links(m0, dirs0), m9, (entries2 -- str)[str0, entries2[str]]) ;mkdir(# entries, md6, # links(ino, dirs0), # subdirs(ino, dirs0), entries) ) = mkdir(m14, md7, m15, m16, entries3), files[entries2[str]] = mkfile(m3, md, # links(entries2[str], dirs0), content0), dirs0[m0] = mkdir(m1 + 1, md5, # links(m0, dirs0), m9, entries2), dirs0[ino] = mkdir(# entries, md6, # links(ino, dirs0), # subdirs(ino, dirs0), entries), files[ino] = mkfile(m8, md3, m10, content), afs-cons(dirs0, files), entries2[str] ∈ files, str ∈ entries2, m0 ∈ dirs0, ROOT_INO ∈ dirs0, ino ∈ dirs0, # links(ino, dirs0) ≤ 1, ino ≠ ROOT_INO ∧ links(ino, dirs0) = ∅ → entries = ∅, name ∈ entries → entries[name] ∈ dirs0 ∨ entries[name] ∈ files, links(m0, dirs0) ≠ ∅ ∨ m0 = ROOT_INO ⊦ m15 = # links(ino, dirs0), str0 ∈ entries2, entries2[str] ∈ dirs0

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: ino = m0

Additional information

Used simplifier rules: ⊦ mkdir(m, md, m0, m1, entries) = mkdir(m2, md0, m3, m4, entries0) ↔ m = m2 ∧ md = md0 ∧ m0 = m3 ∧ m1 = m4 ∧ entries = entries0; (ws) Used formulas: ino = m0