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Proof step 75 from proof abs-paths-eq (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 76:
⊦ ispath(p + q, n, m, dirs, files) ↔ (∃ m0. ispath(p, n, m0, dirs, files) ∧ ispath(q, m0, m, dirs, files))
Proof step 77:
tree[p3].dir?, tree[p3 + p].dir?, abs(tree, ino0)(dirs), ispath(p3, ino0, ino1, dirs, files), ino0 ∈ dirs, p ∈ 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(p, m0, ino1, dirs, files) ⊦ ε = p

Rule argument

spec: afs-base name: path-split seq: ⊦ ispath(p + q, n, m, dirs, files) ↔ (∃ m0. ispath(p, n, m0, dirs, files) ∧ ispath(q, m0, m, dirs, files)) subst: [files, dirs, m, n, q, p] → [files, dirs, ino1, ino0, p, p3] flags: <all paths>

Additional information

Used terms: files, dirs, ino1, ino0, p, p3