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Proof step 42 from proof in-mod (previous step)

tree[p0] = dnode(md0, sdt), p0 ∈ tree, ∀ md2, sdt1, md3, sdt2, q, md4. dnode(md2, sdt1)[p0] = dnode(md3, sdt2) ∧ p0 ∈ dnode(md2, sdt1) → (q ∈ dnode(md2, sdt1)[p0, dnode(md4, sdt2)] ↔ q ∈ dnode(md2, sdt1)) ⊦ q0 ∈ tree[p0, dnode(md1, sdt)] ↔ q0 ∈ tree, p0 = ε, str ∈ sdt1

User applied rule constructor cut

And obtained the following new goals

Proof step 43:
tree[p0] = dnode(md0, sdt), tree = dnode(md, sdt0), p0 ∈ tree, ∀ md2, sdt1, md3, sdt2, q, md4. dnode(md2, sdt1)[p0] = dnode(md3, sdt2) ∧ p0 ∈ dnode(md2, sdt1) → (q ∈ dnode(md2, sdt1)[p0, dnode(md4, sdt2)] ↔ q ∈ dnode(md2, sdt1)) ⊦ q0 ∈ tree[p0, dnode(md1, sdt)] ↔ q0 ∈ tree, p0 = ε, str ∈ sdt1
Proof step 45:
⊦ p ∈ fnode(fid) ↔ p = ε
Proof step 46:
⊦ ε ∈ tree

Rule argument

var: tree

Additional information

Used simplifier rules: fnode-notin: ⊦ p ∈ fnode(fid) ↔ p = ε;%nil-in: ⊦ ε ∈ tree;%directory-tree, entries-tree generated by ∅, dnode, fnode, [ ];;