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Proof step 10 from proof abs-meta-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
tree[p].dir? ↔ m ∈ dirs, ((abs(tree, ino) * D))(dirs), ispath(p, ino, m, dirs, files), p ∈ tree, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ meta(p, tree, abs-files(files)) = meta(ε, m, dirs, files)

Rule argument

Additional information

Used simplifier rules: ⊦ ispath(p, n, m, dirs, files) → ispath(p, n, dirs, files); (s)