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Proof step 19 from proof abs-path-h (previous step)

dirs[m2] = mkdir(m, md, m0, m1, entries), fid = ⌜m2⌝ , D(dirs), ispath(p0, entries[str], dirs, files), m2 ∈ dirs, str ∈ entries, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ tree, ino, D, dirs, files. ((abs(tree, ino) * D))(dirs) ∧ (∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files) ∧ (∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs) → (ispath(p0, ino, dirs, files) ↔ p0 ∈ tree), ∀ ino0. ⌜ino0⌝ = fid → ino0 ∈ files ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
dirs[m2] = mkdir(m, md, m0, m1, entries), D(dirs), ispath(p0, entries[str], dirs, files), str ∈ entries, m2 ∈ dirs, m2 ∈ files, ∀ tree, ino, D, dirs, files. ((abs(tree, ino) * D))(dirs) ∧ (∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files) ∧ (∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs) → (ispath(p0, ino, dirs, files) ↔ p0 ∈ tree), ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ = ⌜m0⌝ ↔ m = m0; (s)