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

p ∈ tree ⊦ tree[p, tree0][p, tree1] = tree[p, tree1]

User applied rule structural induction

And obtained the following new goals

Proof step 4:
⊦ tree[ε, tree0] = tree0
Proof step 5:
str + p0 ∈ tree, ∀ tree, tree0, tree1. p0 ∈ tree → tree[p0, tree0][p0, tree1] = tree[p0, tree1] ⊦ tree[str + p0, tree0][str + p0, tree1] = tree[str + p0, tree1]

Rule argument

indvar: p

Additional information

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