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Proof step 75 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, p ∈ tree, str ∈ entries, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ # links(entries[str], dirs) = 0, ⌜entries[str]⌝ ∈ tree.fids

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 76:
⊦ tree[p] = fnode(fid) ∧ p ∈ tree → fid ∈ tree.fids
Proof step 77:
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

Rule argument

spec: directory-tree-fids name: fids-path seq: ⊦ tree[p] = fnode(fid) ∧ p ∈ tree → fid ∈ tree.fids subst: [p, tree, fid] → [p + str, tree, ⌜entries[str]⌝ ] flags: <all paths>

Additional information

Used terms: p + str, tree, ⌜entries[str]⌝