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Proof step 6 from proof at-del (previous step)

str ' ∈ tree, p0 ∈ tree[str '], q ∈ tree -- (str ' + p0), ∀ tree0, q0. p0 ∈ tree0 ∧ ¬ p0 ⊑ q0 ∧ q0 ∈ tree0 → (tree0 -- p0)[q0] = (q0 ⊑ p0 ⊃ tree0[q0] -- restn(# q0, p0);tree0[q0]) ∨ p0 = ε ⊦ (tree -- (str ' + p0))[q] = (q ⊑ str ' + p0 ⊃ tree[q] -- restn(# q, str ' + p0);tree[q])

User applied rule constructor cut

And obtained the following new goals

Proof step 7:
tree = dnode(md, sdt), q ∈ tree -- (str ' + p0), p0 ∈ tree[str '], str ' ∈ tree, ∀ tree0, q0. p0 ∈ tree0 ∧ ¬ p0 ⊑ q0 ∧ q0 ∈ tree0 → (tree0 -- p0)[q0] = (q0 ⊑ p0 ⊃ tree0[q0] -- restn(# q0, p0);tree0[q0]) ∨ p0 = ε ⊦ (tree -- (str ' + p0))[q] = (q ⊑ str ' + p0 ⊃ tree[q] -- restn(# q, str ' + p0);tree[q])
Proof step 34:
⊦ str ' ∈ fnode(fid) ↔ false

Rule argument

var: tree

Additional information

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