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Proof step 3 from proof split-at (previous step)

p ∈ tree ⊦ tree[p + q] = tree[p][q]

User applied rule structural induction

And obtained the following new goals

Proof step 4:
⊦ tree[ε] = tree
Proof step 5:
str + p0 ∈ tree, ∀ tree, q. p0 ∈ tree → tree[p0 + q] = tree[p0][q] ⊦ tree[(str + p0) + q] = tree[str + p0][q]

Rule argument

indvar: p

Additional information

Used simplifier rules: nil-at: ⊦ tree[ε] = tree;%path freely generated by ε, +;;% ⊦ ε + p = p; (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]