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Proof step 79 from proof inv-rmdir (previous step)

# links(entries0[str], dirs0) = m2 + 1, (ino = m ⊃ mkdir(m3, md3, # links(m, dirs0), m0, entries0 -- str) ;mkdir(# entries1, md0, # links(ino, dirs0), # subdirs(ino, dirs0), entries1) ) = mkdir(m12, md4, m13, m14, entries2), dirs0[m] = mkdir(m3 + 1, md3, # links(m, dirs0), m0 + 1, entries0), dirs0[entries0[str]] = mkdir(m5, md1, m2 + 1, m7, ∅), files[ino] = mkfile(m6, md, m8, content), dirs0[ino] = mkdir(# entries1, md0, # links(ino, dirs0), # subdirs(ino, dirs0), entries1), afs-cons(dirs0, files), name ∈ entries2, ino ∈ dirs0, entries0[str] ∈ dirs0, ROOT_INO ∈ dirs0, m ∈ dirs0, str ∈ entries0, # links(ino, dirs0) ≤ 1, name ∈ entries1 → entries1[name] ∈ dirs0 ∨ entries1[name] ∈ files, ino ≠ ROOT_INO ∧ links(ino, dirs0) = ∅ → entries1 = ∅, links(m, dirs0) ≠ ∅ ∨ m = ROOT_INO ⊦ entries0[str] = ROOT_INO, entries0[str] = m, ino = entries0[str], entries2[name] ∈ dirs0, entries2[name] ∈ files

Heuristic nonsplitting cut applied rule cut formula

And closed the goal!

Rule argument

fma: ino = m

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)%⊦ ino ∈ dirs ∧ dirs[ino] = mkdir(n0, md, n, n1, entries) ∧ str ∈ entries ∧ afs-cons(dirs, files) → (entries[str] ∈ files ↔ ¬ entries[str] ∈ dirs); (ss)% ⊦ str ∈ entries -- str0 ↔ str ≠ str0 ∧ str ∈ entries; (s)% ⊦ str ≠ str0 → (entries -- str0)[str] = entries[str]; (s) Used formulas: ino = m