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Proof step 70 from proof abs-paths-eq (previous step)

tree[p3] = dnode(md, sdt), sdt[str0][p1] = dnode(md0, sdt0), (( ( ( ( ( ((ino1 => mkdir(# entries0, md0, m3, m4, entries0) * abs(sdt0, entries0)) * abs|(sdt[str0], entries[str0], p1, ino1)) * abs|(sdt[str], entries[str], p0, ino1)) * abs(sdt[str][p0], ino1)) * m => mkdir(# entries, md, m1, m2, entries)) * abs(sdt -- str -- str0, entries -- str -- str0)) * abs|(tree, ino0, p3, m)) )(dirs), sdt[str][p0].dir?, ispath(p1, entries[str0], ino1, dirs, files), ispath(p3, ino0, m, dirs, files), ispath(p0, entries[str], ino1, dirs, files), p3 ∈ tree, p1 ∈ sdt[str0], str0 ∈ sdt, ino0 ∈ dirs, str0 ∈ entries, str ∈ entries, str ∈ sdt, p0 ∈ sdt[str], ∀ 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] → [ino1, sdt[str][p0]] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ((_∃ adD * D))(dirs) ↔ (∃ ad. ((adD(ad) * D))(dirs)); (s)% ⊦ (⌜bv⌝ ∧ D) * D0 = (⌜bv⌝ ∧ D * D0); (s)% ⊦ ino => ad * ino => ad0 = false; (s)% ⊦ false * D = false; (s)%sep-ass: ⊦ (D * D0) * D1 = D * D0 * D1; from specification heap-sep[dir-heap]%sep-com: ⊦ D * D0 = D0 * D; from specification heap-sep[dir-heap]% ⊦ (false ∧ D) = false; (s)%and-ass: ⊦ ((D ∧ D0) ∧ D1) = (D ∧ D0 ∧ D1); from specification heap-sep[dir-heap]%and-com: ⊦ (D ∧ D0) = (D0 ∧ D); from specification heap-sep[dir-heap]% ⊦ ¬ false(dirs); (s) Used terms: ino1, sdt[str][p0]