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Proof step 21 from proof abs-paths-eq (previous step)

p2 = str + p0, tree[p3 + p2].dir?, tree[p3 + p].dir?, abs(tree, ino0)(dirs), ispath(p3 + p, ino0, ino1, dirs, files), ispath(p3 + p2, ino0, ino1, dirs, files), ino0 ∈ dirs, p ∈ tree[p3], p2 ∈ tree[p3], p3 ∈ tree, ino1 ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ p = ε, p2 = ε, str = p.head

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
tree[p3 + p].dir?, tree[p3 + str ' + p0].dir?, abs(tree, ino0)(dirs), ispath(p3 + str ' + p0, ino0, ino1, dirs, files), ispath(p3 + p, ino0, ino1, dirs, files), ino1 ∈ dirs, p3 ∈ tree, str ' ∈ tree[p3], p0 ∈ tree[p3][str '], p ∈ tree[p3], ino0 ∈ dirs, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ p = ε, str = p.head

Rule argument

Additional information

Used simplifier rules: ⊦ 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)