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Proof step 31 from proof abs-paths-eq (previous step)

dirs[m] = mkdir(m0, md, m1, m2, entries), tree[p3 + str0 ' + p1].dir?, tree[p3 + str ' + p0].dir?, abs(tree, ino0)(dirs), ispath(p0, entries[str], ino1, dirs, files), ispath(p3, ino0, m, dirs, files), ispath(p1, entries[str0], ino1, dirs, files), str ∈ entries, ino0 ∈ dirs, p1 ∈ tree[p3][str0 '], str0 ' ∈ tree[p3], p0 ∈ tree[p3][str '], str ' ∈ tree[p3], p3 ∈ tree, ino1 ∈ dirs, m ∈ dirs, str0 ∈ entries, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ str = str0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 32:
⊦ p ∈ tree → abs(tree, ino0) = _∃(λ ino1. abs|(tree, ino0, p, ino1) * abs(tree[p], ino1))
Proof step 33:
dirs[m] = mkdir(m0, md, m1, m2, entries), tree[p3 + str ' + p0].dir?, tree[p3 + str0 ' + p1].dir?, (_∃(λ ino1. abs|(tree, ino0, p3, ino1) * abs(tree[p3], ino1)))(dirs), ispath(p1, entries[str0], ino1, dirs, files), ispath(p3, ino0, m, dirs, files), ispath(p0, entries[str], ino1, dirs, files), str0 ∈ entries, m ∈ dirs, ino1 ∈ dirs, str ' ∈ tree[p3], p0 ∈ tree[p3][str '], str0 ' ∈ tree[p3], p1 ∈ tree[p3][str0 '], ino0 ∈ dirs, str ∈ entries, p3 ∈ tree, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ str = str0

Rule argument

spec: <toplevel> name: path-unfold seq: ⊦ p ∈ tree → abs(tree, ino0) = _∃(λ ino1. abs|(tree, ino0, p, ino1) * abs(tree[p], ino1)) subst: [p, ino0, tree] → [p3, ino0, tree] flags: <all paths>

Additional information

Used terms: p3, ino0, tree