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Proof step 102 from proof abs-paths-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 103:
tree[p3] = dnode(md, sdt), (( (abs|(sdt[str], entries[str], p0, m) * abs(sdt[str][p0], m)) * m => mkdir(# entries, md, m1, m2, entries) * abs(sdt -- str, entries -- str) * abs|(tree, ino0, p3, m)) )(dirs), tree[p3 + str ' + p0].dir?, ispath(p0, entries[str], m, dirs, files), ispath(p3, ino0, m, dirs, files), str ∈ entries, p3 ∈ tree, ino0 ∈ dirs, str ∈ sdt, p0 ∈ sdt[str], ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ sdt[str][p0].dir?

Proof step 102 from proof abs-paths-eq (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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