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Proof step 30 from proof abs-paths-eq (previous step)

dirs[m] = mkdir(m0, md, m1, m2, entries), tree[p3 + str ' + p0].dir?, tree[p3 + str0 ' + p1].dir?, abs(tree, ino0)(dirs), ispath(p3, ino0, m, dirs, files), ispath(p0, entries[str], ino1, dirs, files), ino1 ∈ dirs, p3 ∈ tree, str ' ∈ tree[p3], p0 ∈ tree[p3][str '], str0 ' ∈ tree[p3], p1 ∈ tree[p3][str0 '], ino0 ∈ dirs, m ∈ dirs, str ∈ entries, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∃ m0. ispath(p3, ino0, m0, dirs, files) ∧ ispath(str0 ' + p1, m0, ino1, dirs, files) ⊦ str = str0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
dirs[m] = mkdir(m0, md, m1, m2, entries), tree[p3 + str0 ' + p1].dir?, tree[p3 + str ' + p0].dir?, abs(tree, ino0)(dirs), ispath(p0, entries[str], ino1, dirs, files), ispath(p3, ino0, m, dirs, files), ispath(p1, entries[str0], ino1, dirs, files), str ∈ entries, ino0 ∈ dirs, p1 ∈ tree[p3][str0 '], str0 ' ∈ tree[p3], p0 ∈ tree[p3][str '], str ' ∈ tree[p3], p3 ∈ tree, ino1 ∈ dirs, m ∈ dirs, str0 ∈ entries, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ str = str0

Rule argument

Additional information

Used simplifier rules: ⊦ mkdir(m, md, m0, m1, entries).entries = entries; (s)%path-inj: ⊦ ispath(p, n, m0, dirs, files) ∧ ispath(p, n, m1, dirs, files) → m0 = m1; (forward) from specification afs-base%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ 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)