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Proof step 22 from proof abs-meta-h (previous step)

tree[p].dir? ↔ m ∈ dirs, (((abs|(tree, ino, p, m) * abs(tree[p], m)) * D))(dirs), tree[p].dir?, ispath(p, ino, m, dirs, files), p ∈ tree, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ tree[p].meta = meta(ε, m, dirs, files)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
(((abs|(tree, ino, p, m) * abs(tree[p], m)) * D))(dirs), tree[p].dir?, ispath(p, ino, m, dirs, files), p ∈ tree, m ∈ dirs, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ tree[p].meta = dirs[m].meta

Rule argument

Additional information

Used simplifier rules: ⊦ ino ∈ dirs → meta(ε, ino, dirs, files) = dirs[ino].meta; (s)