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Proof step 22 from proof inv-write_inode (previous step)

(ino = m ⊃ mkdir(m5, md, # links(m, dirs), m7, entries);mkdir(m2, md3, m6, m8, entries0)) = mkdir(m9, md4, m10, m11, entries1), dirs[ino] = mkdir(m2, md3, m6, m8, entries0), files[ino] = mkfile(m0, md2, m1, content0), files[m] = mkfile(m3, md0, m4, content), dirs[m] = mkdir(m5, md1, # links(m, dirs), m7, entries), afs-cons(dirs, files), m ∈ dirs, ROOT_INO ∈ dirs, ino ∈ dirs ∧ ino ≠ ROOT_INO ∧ links(ino, dirs) = ∅ → entries0 = ∅, ino ∈ files → ¬ ino ∈ dirs ∧ m1 = # links(ino, dirs), ino ∈ dirs ∧ name ∈ entries0 → entries0[name] ∈ dirs ∨ entries0[name] ∈ files, ino ∈ dirs → m6 = # links(ino, dirs) ∧ # links(ino, dirs) ≤ 1 ∧ m2 = # entries0 ∧ m8 = # subdirs(ino, dirs) ⊦ (ino ∈ dirs ∧ ino ≠ ROOT_INO ∧ links(ino, dirs) = ∅ → entries1 = ∅) ∧ (ino ∈ dirs → m10 = # links(ino, dirs) ∧ # links(ino, dirs) ≤ 1 ∧ m9 = # entries1 ∧ m11 = # subdirs(ino, dirs)) ∧ (ino ∈ dirs ∧ name ∈ entries1 → entries1[name] ∈ dirs ∨ entries1[name] ∈ files)

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

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)% ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs → ¬ ino ∈ files; (s)%afs-cons-dir-nlink: ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs ∧ dirs[ino] = mkdir(n0, md, n, n1, entries) → # links(ino, dirs) = n; (forward) from specification afs-base%⊦ mkfile(m, md, m0, content) = mkfile(m1, md0, m2, content0) ↔ m = m1 ∧ md = md0 ∧ m0 = m2 ∧ content = content0; (ws)