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Proof step 17 from proof abs-tree (previous step)

fid = ⌜m⌝ , (((m => mkdir(# entries, md, m0, m1, entries) * abs(sdt, entries)) * D0))(dirs), D1(dirs), ∀ ino0. ⌜ino0⌝ ∈ sdt.fids → ino0 ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino0. ⌜ino0⌝ = fid → ino0 ∈ files ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
(((m => mkdir(# entries, md, m0, m1, entries) * abs(sdt, entries)) * D0))(dirs), D1(dirs), m ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino0. ⌜ino0⌝ ∈ sdt.fids → ino0 ∈ files ⊦

Rule argument

Additional information

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