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Proof step 5 from proof split-in (previous step)

∀ 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 6:
∀ q, tree. p0 + q ∈ tree ↔ p0 ∈ tree ∧ q ∈ tree[p0] ⊦ (str ' + p0) + q ∈ tree ↔ str ' + p0 ∈ tree ∧ q ∈ tree[str ' + p0]

Rule argument

Additional information

Used simplifier rules: a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ str + p = str ' + p; (s)