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Proof step 6 from proof in-del (previous step)

tree = dnode(md, sdt), p0 ∈ tree[str '], str ' ∈ tree, ∀ tree, q. p0 ∈ tree → (q ∈ tree -- p0 ↔ ¬ p0 ⊑ q ∧ q ∈ tree) ∨ p0 = ε ⊦ q ∈ tree -- (str ' + p0) ↔ ¬ str ' + p0 ⊑ q ∧ q ∈ tree

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ str ' ∈ dnode(md, sdt) ↔ str ∈ sdt
Proof step 8:
str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str]
Proof step 9:
str ∈ sdt, p0 ∈ sdt[str], ∀ tree, q. p0 ∈ tree → (q ∈ tree -- p0 ↔ ¬ p0 ⊑ q ∧ q ∈ tree) ∨ p0 = ε ⊦ q ∈ dnode(md, sdt) -- (str ' + p0) ↔ ¬ str ' + p0 ⊑ q ∧ q ∈ dnode(md, sdt)

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];