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Proof step 59 from proof abs-paths-eq (previous step)

tree[p3] = dnode(md, sdt), (( (abs|(sdt[str0], entries[str0], p1, ino1) * abs(sdt[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), 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), p1 ∈ sdt[str0], str0 ∈ sdt, ino0 ∈ dirs, str0 ∈ entries, str ∈ entries, ino1 ∈ dirs, str ∈ sdt, p0 ∈ sdt[str], p3 ∈ tree, ∀ 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 60:
tree[p3] = dnode(md, sdt), (( (abs|(sdt[str0], entries[str0], p1, ino1) * abs(sdt[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), tree[p3 + str0 ' + p1].dir?, sdt[str][p0].dir?, ispath(p1, entries[str0], ino1, dirs, files), ispath(p3, ino0, m, dirs, files), ispath(p0, entries[str], ino1, dirs, files), p1 ∈ sdt[str0], str0 ∈ sdt, ino0 ∈ dirs, str0 ∈ entries, str ∈ entries, ino1 ∈ dirs, str ∈ sdt, p0 ∈ sdt[str], p3 ∈ tree, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ str = str0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ str ∈ sdt → dnode(md, sdt)[str ' + p] = sdt[str][p]; (s)