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Proof step 20 from proof abs-dir-h (previous step)

((ino => mkdir(# entries, md, m0, m1, entries) * (⌜str ∈ entries⌝ ∧ abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str])) * D))(dirs), ispath(p0, entries[str], ino0, dirs, files), str ∈ entries, str ∈ sdt, ∀ ino1. ⌜ino1⌝ ∈ sdt.fids → ino1 ∈ 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 ⊦ sdt[str][p0].dir? ↔ ino0 ∈ dirs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
((ino => mkdir(# entries, md, m0, m1, entries) * (abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str])) * D))(dirs), ispath(p0, entries[str], ino0, dirs, files), str ∈ sdt, str ∈ entries, ∀ 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⌝ ∈ sdt.fids → ino1 ∈ files ⊦ sdt[str][p0].dir? ↔ ino0 ∈ dirs

Rule argument

Additional information

Used simplifier rules: and-com: ⊦ (D ∧ D0) = (D0 ∧ D); from specification heap-sep[dir-heap]%and-ass: ⊦ ((D ∧ D0) ∧ D1) = (D ∧ D0 ∧ D1); from specification heap-sep[dir-heap]% ⊦ (true ∧ D) = D; (s)% ⊦ ⌜true⌝ = true; (s)