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Proof step 2 from proof split-in (previous step)

⊦ p + q ∈ tree ↔ p ∈ tree ∧ q ∈ tree[p]

User applied rule structural induction

And obtained the following new goals

Proof step 3:
⊦ ε ∈ tree
Proof step 4:
⊦ tree[ε] = tree
Proof step 5:
∀ q, tree. p0 + q ∈ tree ↔ p0 ∈ tree ∧ q ∈ tree[p0] ⊦ (str + p0) + q ∈ tree ↔ str + p0 ∈ tree ∧ q ∈ tree[str + p0]

Rule argument

indvar: p

Additional information

Used simplifier rules: nil-in: ⊦ ε ∈ tree;%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]