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Proof step 4 from proof again-put (previous step)

str + p0 ∈ tree, ∀ tree. p0 ∈ tree → tree[p0, tree[p0]] = tree ∨ p0 = ε ⊦ str + p0 = ε, tree[str + p0, tree[str + p0]] = tree

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ p + q ∈ tree ↔ p ∈ tree ∧ q ∈ tree[p]
Proof step 6:
⊦ p ∈ tree → tree[p + q] = tree[p][q]
Proof step 7:
⊦ p ∈ tree → tree[p + q, tree0] = tree[p, tree[p][q, tree0]]
Proof step 8:
str ' ∈ tree, p0 ∈ tree[str '], ∀ tree. p0 ∈ tree → tree[p0, tree[p0]] = tree ∨ p0 = ε ⊦ tree[str ', tree[str '][p0, tree[str '][p0]]] = tree

Rule argument

Additional information

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