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Proof step 77 from proof abs-nlinks-h (previous step)

tree[p] = dnode(md, sdt), ((m => mkdir(# entries, md, m1, m2, entries) * (abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str])) * abs|(tree, ino0, p, m)))(dirs), ispath(p, ino0, m, dirs, files), str ∈ entries, p ∈ tree, ino0 ∈ dirs, entries[str] ∈ files, str ∈ sdt, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ # links(entries[str], dirs) = 0, ⌜entries[str]⌝ ∈ tree.fids, tree[p + str] = fnode(⌜entries[str]⌝ ) ∧ p + str ∈ tree

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 78:
tree[p] = dnode(md, sdt), ((m => mkdir(# entries, md, m1, m2, entries) * (abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str])) * abs|(tree, ino0, p, m)))(dirs), ispath(p, ino0, m, dirs, files), str ∈ entries, p ∈ tree, ino0 ∈ dirs, entries[str] ∈ files, str ∈ sdt, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ # links(entries[str], dirs) = 0, tree[p + str '] = fnode(⌜entries[str]⌝ ), ⌜entries[str]⌝ ∈ tree.fids

Rule argument

Additional information

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