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Proof step 91 from proof abs-paths-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 92:
⊦ str ∈ sdi ∧ ((abs(sdt, sdi) * D))(dirs) → str ∈ sdt
Proof step 93:
tree[p3] = dnode(md, sdt), ((m => mkdir(# entries, md, m1, m2, entries) * abs(sdt, entries) * abs|(tree, ino0, p3, m)))(dirs), tree[p3 + str ' + p0].dir?, ispath(p3, ino0, m, dirs, files), ispath(p0, entries[str], m, dirs, files), ino0 ∈ dirs, p3 ∈ tree, str ∈ entries, str ∈ sdt, p0 ∈ sdt[str], ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦

Rule argument

Additional information

Used simplifier rules: entries-dom1: ⊦ str ∈ sdi ∧ ((abs(sdt, sdi) * D))(dirs) → str ∈ sdt;%str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str]; (s)% ⊦ str ' ∈ dnode(md, sdt) ↔ str ∈ sdt; (s)% ⊦ dnode(md, sdt).dir?; (s)%sep-com: ⊦ D * D0 = D0 * D; from specification heap-sep[dir-heap]%sep-ass: ⊦ (D * D0) * D1 = D * D0 * D1; from specification heap-sep[dir-heap]