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Proof step 3 from proof abs-fold-new-p (previous step)

tree[p] = dnode(md, sdt), ((abs|(tree, ROOT_INO, p, ino0) * ino0 => mkdir(# entries + 1, md, m0, m1, entries[str, ino1]) * abs(sdt, entries) * D))(dirs), p ∈ tree, D ⊆ abs(tree0, ino1) ⊦ abs(tree[p + str ', tree0], ROOT_INO)(dirs), str ∈ entries

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ p ∈ tree → tree[p + q, tree0] = tree[p, tree[p][q, tree0]]
Proof step 5:
tree[p] = dnode(md, sdt), ((abs|(tree, ROOT_INO, p, ino0) * ino0 => mkdir(# entries + 1, md, m0, m1, entries[str, ino1]) * abs(sdt, entries) * D))(dirs), p ∈ tree, D ⊆ abs(tree0, ino1) ⊦ abs(tree[p, tree[p][str ', tree0]], ROOT_INO)(dirs), str ∈ entries

Rule argument

spec: directory-tree name: split-put seq: ⊦ p ∈ tree → tree[p + q, tree0] = tree[p, tree[p][q, tree0]] subst: [tree0, q, p, tree] → [tree0, str ', p, tree] flags: <all paths>

Additional information

Used terms: tree0, str ', p, tree