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Proof step 98 from proof abs-paths-eq (previous step)

tree[p3] = dnode(md, sdt), (( m => mkdir(# entries, md, m1, m2, entries) * (abs(sdt -- str, entries -- str) * _∃(λ ino1. abs|(sdt[str], entries[str], p0, ino1) * abs(sdt[str][p0], ino1))) * 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 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 99:
⊦ p ∈ tree ∧ ((abs|(tree, ino, p, ino0) * D))(dirs) ∧ ispath(p, ino, ino1, dirs, files) → ino0 = ino1
Proof step 100:
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 ⊦

Rule argument

Additional information

Used simplifier rules: absc-path: ⊦ p ∈ tree ∧ ((abs|(tree, ino, p, ino0) * D))(dirs) ∧ ispath(p, ino, ino1, dirs, files) → ino0 = ino1;%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]% ⊦ ((_∃ nD * D))(dirs) ↔ (∃ ino. ((nD(ino) * D))(dirs)); (s)