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Proof step 10 from proof path-split-wand (previous step)

abs(tree, ino0)(dirs), str ' ∈ tree, p0 ∈ tree[str '], ∀ tree, ino0, dirs. abs(tree, ino0)(dirs) ∧ p0 ∈ tree → (∃ ino. ((abs(tree[p0], ino) * (abs(tree[p0], ino) -* abs(tree, ino0))))(dirs)) ⊦ ∃ ino. ((abs(tree[str ' + p0], ino) * (abs(tree[str ' + p0], ino) -* abs(tree, ino0))))(dirs)

User applied rule constructor cut

And obtained the following new goals

Proof step 11:
tree = dnode(md, sdt), abs(tree, ino0)(dirs), p0 ∈ tree[str '], str ' ∈ tree, ∀ tree, ino0, dirs. abs(tree, ino0)(dirs) ∧ p0 ∈ tree → (∃ ino. ((abs(tree[p0], ino) * (abs(tree[p0], ino) -* abs(tree, ino0))))(dirs)) ⊦ ∃ ino. ((abs(tree[str ' + p0], ino) * (abs(tree[str ' + p0], ino) -* abs(tree, ino0))))(dirs)

Rule argument

var: tree

Additional information

Used simplifier rules: directory-tree, entries-tree generated by ∅, dnode, fnode, [ ];;% ⊦ str ' ∈ fnode(fid) ↔ false; (s)