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Proof step 5 from proof in-del (previous step)

str ' ∈ tree, p0 ∈ tree[str '], ∀ tree, q. p0 ∈ tree → (q ∈ tree -- p0 ↔ ¬ p0 ⊑ q ∧ q ∈ tree) ∨ p0 = ε ⊦ q ∈ tree -- (str ' + p0) ↔ ¬ str ' + p0 ⊑ q ∧ q ∈ tree

User applied rule constructor cut

And obtained the following new goals

Proof step 6:
tree = dnode(md, sdt), p0 ∈ tree[str '], str ' ∈ tree, ∀ tree, q. p0 ∈ tree → (q ∈ tree -- p0 ↔ ¬ p0 ⊑ q ∧ q ∈ tree) ∨ p0 = ε ⊦ q ∈ tree -- (str ' + p0) ↔ ¬ str ' + p0 ⊑ q ∧ q ∈ tree
Proof step 28:
⊦ p ∈ fnode(fid) ↔ p = ε
Proof step 29:
⊦ ε ∈ tree
Proof step 30:
⊦ str ' ∈ fnode(fid) ↔ false

Rule argument

var: tree

Additional information

Used simplifier rules: fnode-notin: ⊦ p ∈ fnode(fid) ↔ p = ε;%nil-in: ⊦ ε ∈ tree;%one-inf: ⊦ str ' ∈ fnode(fid) ↔ false;%directory-tree, entries-tree generated by ∅, dnode, fnode, [ ];;% ⊦ p ⊑ ε ↔ p = ε; (s)% ⊦ str ' + p ≠ ε; (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]