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Proof step 8 from proof split-at (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ str ' ∈ dnode(md, sdt) ↔ str ∈ sdt
Proof step 10:
⊦ str ∈ sdt → dnode(md, sdt)[str ' + p] = sdt[str][p]
Proof step 11:
str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str]
Proof step 12:
str ∈ sdt, p0 ∈ sdt[str], ∀ tree, q. p0 ∈ tree → tree[p0 + q] = tree[p0][q] ⊦ sdt[str][p0 + q] = sdt[str][p0][q]

Rule argument

Additional information

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