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Proof step 80 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 ∈ sdt, entries[str] ∈ files, ino0 ∈ dirs, str ∈ entries, p ∈ tree, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ # links(entries[str], dirs) = 0, tree[p][str '] = fnode(⌜entries[str]⌝ ), ⌜entries[str]⌝ ∈ tree.fids

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 81:
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 ∈ sdt, entries[str] ∈ files, ino0 ∈ dirs, str ∈ entries, p ∈ tree, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ # links(entries[str], dirs) = 0, sdt[str] = fnode(⌜entries[str]⌝ ), ⌜entries[str]⌝ ∈ tree.fids

Rule argument

Additional information

Used simplifier rules: str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str]; (s)