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Proof step 20 from proof abs-paths-eq (previous step)

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], ino0 ∈ dirs, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ p2.head = p.head, p = ε, p2 = ε

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 21:
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

Rule argument

spec: list-data, inst: path name: elim-list-c seq: p ≠ ε ⊦ str = p.head ∧ p0 = p.tail ↔ p = str + p0 subst: [p, str, p0] → [p2, str, p0]

Additional information

Used simplifier rules: p ≠ ε ⊦ str = p.head ∧ p0 = p.tail ↔ p = str + p0; (elim) Used terms: p2, str, p0