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Proof step 95 from proof abs-paths-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 96:
tree[p3] = dnode(md, sdt), ((m => mkdir(# entries, md, m1, m2, entries) * (abs(sdt -- str, entries -- str) * 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), str ∈ sdt, ino0 ∈ dirs, p3 ∈ tree, str ∈ entries, p0 ∈ sdt[str], ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦

Rule argument

Additional information

Used simplifier rules: and-com: ⊦ (D ∧ D0) = (D0 ∧ D); from specification heap-sep[dir-heap]%and-ass: ⊦ ((D ∧ D0) ∧ D1) = (D ∧ D0 ∧ D1); from specification heap-sep[dir-heap]% ⊦ (true ∧ D) = D; (s)% ⊦ ⌜true⌝ = true; (s)