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Proof step 17 from proof abs-paths-eq (previous step)

tree[p0].dir?, tree[p1].dir?, abs(tree, ino0)(dirs), ispath(p1, ino0, ino1, dirs, files), ispath(p0, ino0, ino1, dirs, files), p1 ∈ tree, p0 ∈ tree, ino1 ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∃ p, p5, p4. p0 = p + p5 ∧ p1 = p + p4 ∧ (p5 ≠ ε ∧ p4 ≠ ε → p5.head ≠ p4.head) ⊦ p0 = p1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
tree[p3 + p].dir?, tree[p3 + p2].dir?, abs(tree, ino0)(dirs), ispath(p3 + p2, ino0, ino1, dirs, files), ispath(p3 + p, ino0, ino1, dirs, files), ino1 ∈ dirs, p3 ∈ tree, p2 ∈ tree[p3], p ∈ tree[p3], p2 ≠ ε ∧ p ≠ ε → p2.head ≠ p.head, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ p2 = p

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]% ⊦ p + p0 = p + p1 ↔ p0 = p1; (s)