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Proof step 39 from proof inv-evict (previous step)

# links(m1, dirs) = 0, dirs[ino] = mkdir(m3, md1, m4, m5, entries), files[ino] = mkfile(m, md, m0, content), files[m1] = mkfile(m2, md0, 0, content0), afs-cons(dirs, files), m1 ∈ files, ROOT_INO ∈ dirs, ino ∈ dirs ∧ name ∈ entries → entries[name] ∈ dirs ∨ entries[name] ∈ files, ino ∈ files → ¬ ino ∈ dirs ∧ m0 = # links(ino, dirs), ino ∈ dirs ∧ ino ≠ ROOT_INO ∧ links(ino, dirs) = ∅ → entries = ∅, ino ∈ dirs → m4 = # links(ino, dirs) ∧ # links(ino, dirs) ≤ 1 ∧ m3 = # entries ∧ m5 = # subdirs(ino, dirs) ⊦ m1 ∈ dirs, (ino ≠ m1 ∧ ino ∈ files → ¬ ino ∈ dirs ∧ m0 = # links(ino, dirs)) ∧ (ino ∈ dirs ∧ name ∈ entries → entries[name] ∈ dirs ∨ entries[name] ≠ m1 ∧ entries[name] ∈ files)

Heuristic pl case distinction applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: ⊦ 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% ⊦ # ∅ = 0; (s)% ⊦ ¬ str ∈ ∅; (s)%size-zero-02: ⊦ # ls = 0 → ls = ∅; (forward) from specification set-basic[alink-set]%afs-cons-file-nlink: ⊦ afs-cons(dirs, files) ∧ ino ∈ files ∧ files[ino] = mkfile(m, md, n, content) → # links(ino, dirs) = n; (forward) from specification afs-base%⊦ p_ino ∈ dirs ∧ dirs[p_ino] = mkdir(m, md, n, n0, entries) ∧ str ∈ entries → links(entries[str], dirs) ≠ ∅; (s)