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Proof step 26 from proof abs-paths-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
tree[p3 + str ' + p0].dir?, tree[p3 + str0 ' + p1].dir?, abs(tree, ino0)(dirs), ispath(p3 + str0 ' + p1, ino0, ino1, dirs, files), ispath(p3, ino0, m, dirs, files), ispath(p0, dirs[m].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 ∈ dirs[m].entries, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ str = str0

Rule argument

Additional information

Used simplifier rules: 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)