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Proof step 44 from proof abs-unshared-h (previous step)

tree1[p1] = fnode(⌜m⌝ ), (( ( ((m => mkdir(# entries, tree0[p0].meta, m0, m1, entries) * abs(tree0[p0].entries, entries)) * abs|(tree1, ino1, p1, m)) * abs|(tree0, ino0, p0, m)) * D) )(dirs), tree0[p0].dir?, ispath(p1, ino1, m, dirs, files), ispath(p0, ino0, m, dirs, files), p1 ∈ tree1, p0 ∈ tree0, ⌜m⌝ ∈ tree1.fids → m ∈ files, ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files, ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 45:
tree1[p1] = fnode(⌜m⌝ ), (( ( ((m => mkdir(# entries, tree0[p0].meta, m0, m1, entries) * abs(tree0[p0].entries, entries)) * abs|(tree1, ino1, p1, m)) * abs|(tree0, ino0, p0, m)) * D) )(dirs), tree0[p0].dir?, ispath(p0, ino0, m, dirs, files), ispath(p1, ino1, m, dirs, files), p0 ∈ tree0, p1 ∈ tree1, m ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files, ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ tree[p] = fnode(fid) ∧ p ∈ tree → fid ∈ tree.fids; (ss)