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Proof step 69 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?, sdt[str0][p1].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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 70:
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

Rule argument

Additional information

Used simplifier rules: ⊦ dnode(md, sdt).dir?; (s)