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Proof step 83 from proof abs-paths-eq (previous step)

p = str + p0, ((abs|(tree, ino0, p3, m) * abs(tree[p3], m)))(dirs), tree[p3].dir?, tree[p3 + p].dir?, ispath(p, m, m, dirs, files), ispath(p3, ino0, m, dirs, files), p3 ∈ tree, ino0 ∈ dirs, p ∈ tree[p3], m ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ ε = p

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 84:
((abs|(tree, ino0, p3, m) * abs(tree[p3], m)))(dirs), tree[p3].dir?, tree[p3 + str ' + p0].dir?, ispath(p3, ino0, m, dirs, files), ispath(p0, dirs[m].entries[str], m, dirs, files), m ∈ dirs, str ' ∈ tree[p3], p0 ∈ tree[p3][str '], ino0 ∈ dirs, p3 ∈ tree, str ∈ dirs[m].entries, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ valid-dir-ino(ino, dirs, files) ↔ ino ∈ dirs; (s)%⊦ ispath(str ' + p, n, m, dirs, files) ↔ valid-dir-ino(n, dirs, files) ∧ str ∈ dirs[n].entries ∧ ispath(p, dirs[n].entries[str], m, dirs, files); (s)% ⊦ p + q ∈ tree ↔ p ∈ tree ∧ q ∈ tree[p]; (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ str ' + p ≠ ε; (s)% ⊦ str + p = str ' + p; (s)