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Proof step 12 from proof in-put (previous step)

q = str0 + q0 ⊦ q ∈ dnode(md, sdt[str, tree0]) ↔ (str ' ⊑ q ⊃ q.tail ∈ tree0;q ∈ dnode(md, sdt))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
⊦ str ' + p ∈ dnode(md, sdt) ↔ str ∈ sdt ∧ p ∈ sdt[str]
Proof step 14:
⊦ str0 ∈ sdt[str, tree0] ∧ q0 ∈ sdt[str, tree0][str0] ↔ (str = str0 ⊃ q0 ∈ tree0;str0 ∈ sdt ∧ q0 ∈ sdt[str0])

Rule argument

Additional information

Used simplifier rules: in-dnode: ⊦ str ' + p ∈ dnode(md, sdt) ↔ str ∈ sdt ∧ p ∈ sdt[str];% ⊦ (str ' + p).tail = p; (s)% ⊦ str ' ⊑ str0 ' + p ↔ str = str0; (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ str + p = str ' + p; (s)