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Proof step 7 from proof split-in (previous step)

tree = dnode(md, sdt), ∀ q, tree. p0 + q ∈ tree ↔ p0 ∈ tree ∧ q ∈ tree[p0] ⊦ (str ' + p0) + q ∈ tree ↔ str ' + p0 ∈ tree ∧ q ∈ tree[str ' + p0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ str ' + p ∈ dnode(md, sdt) ↔ str ∈ sdt ∧ p ∈ sdt[str]
Proof step 9:
⊦ str ∈ sdt → dnode(md, sdt)[str ' + p] = sdt[str][p]
Proof step 10:
str ∈ sdt, ∀ q, tree. p0 + q ∈ tree ↔ p0 ∈ tree ∧ q ∈ tree[p0] ⊦ p0 + q ∈ sdt[str] ↔ p0 ∈ sdt[str] ∧ q ∈ sdt[str][p0]

Rule argument

Additional information

Used simplifier rules: in-dnode: ⊦ str ' + p ∈ dnode(md, sdt) ↔ str ∈ sdt ∧ p ∈ sdt[str];%at-dnode: ⊦ str ∈ sdt → dnode(md, sdt)[str ' + p] = sdt[str][p];%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]