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Proof step 11 from proof in-del (previous step)

q = str0 + q0, p0 ∈ sdt[str], str ∈ sdt, ∀ tree, q. p0 ∈ tree → (q ∈ tree -- p0 ↔ ¬ p0 ⊑ q ∧ q ∈ tree) ∨ p0 = ε ⊦ q ∈ dnode(md, sdt) -- (str ' + p0) ↔ ¬ str ' + p0 ⊑ q ∧ q ∈ dnode(md, sdt)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
⊦ p + q ∈ tree ↔ p ∈ tree ∧ q ∈ tree[p]
Proof step 13:
⊦ str ' + p ∈ dnode(md, sdt) ↔ str ∈ sdt ∧ p ∈ sdt[str]
Proof step 14:
str ∈ sdt, p0 ∈ sdt[str], ∀ tree, q. p0 ∈ tree → (q ∈ tree -- p0 ↔ ¬ p0 ⊑ q ∧ q ∈ tree) ∨ p0 = ε ⊦ str0 ' ∈ dnode(md, sdt) -- (str ' + p0) ∧ q0 ∈ (dnode(md, sdt) -- (str ' + p0))[str0 '] ↔ ¬ str ' + p0 ⊑ str0 ' + q0 ∧ str0 ∈ sdt ∧ q0 ∈ sdt[str0]

Rule argument

Additional information

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