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Proof step 10 from proof path-unfold (previous step)

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

User applied rule constructor cut

And obtained the following new goals

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

Rule argument

var: tree

Additional information

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