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Proof step 35 from proof abs-paths-eq (previous step)

dirs[m] = mkdir(m0, md, m1, m2, entries), ((abs|(tree, ino0, p3, m) * abs(tree[p3], m)))(dirs), tree[p3 + str ' + p0].dir?, tree[p3 + str0 ' + p1].dir?, ispath(p1, entries[str0], ino1, dirs, files), ispath(p3, ino0, m, dirs, files), ispath(p0, entries[str], ino1, dirs, files), str0 ∈ entries, m ∈ dirs, ino1 ∈ dirs, str ' ∈ tree[p3], p0 ∈ tree[p3][str '], str0 ' ∈ tree[p3], p1 ∈ tree[p3][str0 '], ino0 ∈ dirs, str ∈ entries, p3 ∈ tree, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ str = str0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: abs-is-dnode seq: ⊦ tree.dir? → abs(tree, ino) = _∃ (λ ad. ino => ad * abs(tree.entries, ad.entries) ∧ ⌜ad.meta = tree.meta⌝ ∧ ⌜ad.size = # ad.entries⌝ ) subst: [ino, tree] → [m, tree[p3]] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ¬ tree.dir? → (p ∈ tree ↔ p = ε); (s)% ⊦ str ' ≠ ε; (s) Used terms: m, tree[p3]