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Proof step 9 from proof again-put (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
⊦ dnode(md, sdt)[str ', tree0] = dnode(md, sdt[str, tree0])
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. p0 ∈ tree → tree[p0, tree[p0]] = tree ∨ p0 = ε ⊦ dnode(md, sdt[str, sdt[str][p0, sdt[str][p0]]]) = dnode(md, sdt)

Rule argument

Additional information

Used simplifier rules: put-here: ⊦ dnode(md, sdt)[str ', tree0] = dnode(md, sdt[str, tree0]);%one-ind: ⊦ str ' ∈ dnode(md, sdt) ↔ str ∈ sdt;%one-at: str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str];