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Proof step 15 from proof abs-paths-eq (previous step)

tree[p1].dir?, tree[p0].dir?, abs(tree, ino0)(dirs), ispath(p0, ino0, ino1, dirs, files), ispath(p1, ino0, ino1, dirs, files), ino1 ∈ dirs, p0 ∈ tree, p1 ∈ tree, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ p0 = p1

User applied rule apply lemma

And obtained the following new goals

Proof step 16:
⊦ ∃ p, p0, p1. p2 = p + p0 ∧ p3 = p + p1 ∧ (p0 ≠ ε ∧ p1 ≠ ε → p0.head ≠ p1.head)
Proof step 17:
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

Rule argument

spec: list, inst: path name: longest-common-prefix seq: ⊦ ∃ p, p0, p1. p2 = p + p0 ∧ p3 = p + p1 ∧ (p0 ≠ ε ∧ p1 ≠ ε → p0.head ≠ p1.head) subst: [p2, p3] → [p0, p1]

Additional information

Used terms: p0, p1