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

str + p0 ∈ 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 6:
⊦ p + q ∈ tree ↔ p ∈ tree ∧ q ∈ tree[p]
Proof step 7:
str ' ∈ tree, p0 ∈ tree[str '], ∀ tree, q. p0 ∈ tree → tree[p0 + q] = tree[p0][q] ⊦ tree[(str ' + p0) + q] = tree[str ' + p0][q]

Rule argument

Additional information

Used simplifier rules: split-in: ⊦ p + q ∈ tree ↔ p ∈ tree ∧ q ∈ tree[p];%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ str + p = str ' + p; (s)