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Proof step 6 from proof same-in (previous step)

str + p0 ∈ tree, ∀ tree, tree0. p0 ∈ tree → p0 ∈ tree[p0, tree0] ⊦ str + p0 ∈ tree[str + p0, tree0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ p + q ∈ tree ↔ p ∈ tree ∧ q ∈ tree[p]
Proof step 8:
⊦ p ∈ tree → tree[p + q, tree0] = tree[p, tree[p][q, tree0]]
Proof step 9:
⊦ p ∈ tree → tree[p, tree0][p] = tree0
Proof step 10:
str ' ∈ tree, p0 ∈ tree[str '], ∀ tree, tree0. p0 ∈ tree → p0 ∈ tree[p0, tree0] ⊦ str ' ∈ tree[str ', tree[str '][p0, tree0]] ∧ p0 ∈ tree[str '][p0, tree0]

Rule argument

Additional information

Used simplifier rules: split-in: ⊦ p + q ∈ tree ↔ p ∈ tree ∧ q ∈ tree[p];%split-put: ⊦ p ∈ tree → tree[p + q, tree0] = tree[p, tree[p][q, tree0]];%same-at: ⊦ p ∈ tree → tree[p, tree0][p] = tree0;%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ str + p = str ' + p; (s)