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Proof step 3 from proof same-in (previous step)

p ∈ tree ⊦ p ∈ tree[p, tree0]

User applied rule structural induction

And obtained the following new goals

Proof step 4:
⊦ ε ∈ tree
Proof step 5:
⊦ tree[ε, tree0] = tree0
Proof step 6:
str + p0 ∈ tree, ∀ tree, tree0. p0 ∈ tree → p0 ∈ tree[p0, tree0] ⊦ str + p0 ∈ tree[str + p0, tree0]

Rule argument

indvar: p

Additional information

Used simplifier rules: nil-in: ⊦ ε ∈ tree;%nil-put: ⊦ tree[ε, tree0] = tree0;%path freely generated by ε, +;;