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Proof step 10 from proof split-put (previous step)

tree = dnode(md, sdt), p0 ∈ tree[str '], str ' ∈ tree, ∀ tree, q, tree0. p0 ∈ tree → tree[p0 + q, tree0] = tree[p0, tree[p0][q, tree0]] ⊦ tree[(str ' + p0) + q, tree0] = tree[str ' + p0, tree[str '][p0][q, tree0]]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
⊦ str ' ∈ dnode(md, sdt) ↔ str ∈ sdt
Proof step 12:
str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str]
Proof step 13:
str ∈ sdt, p0 ∈ sdt[str], ∀ tree, q, tree0. p0 ∈ tree → tree[p0 + q, tree0] = tree[p0, tree[p0][q, tree0]] ⊦ dnode(md, sdt)[(str ' + p0) + q, tree0] = dnode(md, sdt)[str ' + p0, sdt[str][p0][q, tree0]]

Rule argument

Additional information

Used simplifier rules: one-ind: ⊦ str ' ∈ dnode(md, sdt) ↔ str ∈ sdt;%one-at: str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str];