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Proof step 27 from proof abs-dir-h (previous step)

dirs[m2] = mkdir(m, md, m0, m1, entries), fid = ⌜m2⌝ , D(dirs), ispath(p0, entries[str], ino0, dirs, files), str ∈ entries, m2 ∈ dirs, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ tree, ino, D, dirs, ino0, files. ((abs(tree, ino) * D))(dirs) ∧ ispath(p0, ino, ino0, dirs, files) ∧ (∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files) ∧ (∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs) → (tree[p0].dir? ↔ ino0 ∈ dirs), ∀ ino1. ⌜ino1⌝ = fid → ino1 ∈ files ⊦ fnode(fid)[str ' + p0].dir? ↔ ino0 ∈ dirs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
dirs[m2] = mkdir(m, md, m0, m1, entries), D(dirs), ispath(p0, entries[str], ino0, dirs, files), m2 ∈ dirs, str ∈ entries, m2 ∈ files, ∀ tree, ino, D, dirs, ino0, files. ((abs(tree, ino) * D))(dirs) ∧ ispath(p0, ino, ino0, dirs, files) ∧ (∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files) ∧ (∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs) → (tree[p0].dir? ↔ ino0 ∈ dirs), ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ fnode(⌜m2⌝ )[str ' + p0].dir? ↔ ino0 ∈ dirs

Rule argument

Additional information

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