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Proof step 33 from proof in-put (previous step)

sdt0[str0] = dnode(md, sdt), str1 = str0, str0 ∈ sdt0, ∀ md1, sdt1, md2, sdt2, q, str2, tree. dnode(md1, sdt1) = dnode(md2, sdt2) → (q ∈ dnode(md2, sdt2[str2, tree]) ↔ (str2 ' ⊑ q ⊃ q.tail ∈ tree;q ∈ dnode(md1, sdt1))) ⊦ str1 ∈ sdt0[str0, dnode(md, sdt[str, tree0])] ∧ q0 ∈ dnode(md0, sdt0[str0, dnode(md, sdt[str, tree0])])[str1 '] ↔ (str0 ' + str ' ⊑ str1 ' + q0 ⊃ q0.tail ∈ tree0;str1 ∈ sdt0 ∧ q0 ∈ sdt0[str1])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str]
Proof step 35:
sdt0[str0] = dnode(md, sdt), str0 ∈ sdt0, ∀ md1, sdt1, md2, sdt2, q, str2, tree. dnode(md1, sdt1) = dnode(md2, sdt2) → (q ∈ dnode(md2, sdt2[str2, tree]) ↔ (str2 ' ⊑ q ⊃ q.tail ∈ tree;q ∈ dnode(md1, sdt1))) ⊦ q0 ∈ dnode(md, sdt[str, tree0]) ↔ (str ' ⊑ q0 ⊃ q0.tail ∈ tree0;q0 ∈ dnode(md, sdt))

Rule argument

Additional information

Used simplifier rules: one-at: str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str];%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ str ' + p ⊑ str ' + p0 ↔ p ⊑ p0; (s)% ⊦ sdt[str, tree][str] = tree; (s)% ⊦ str ∈ sdt[str0, tree] ↔ str = str0 ∨ str ∈ sdt; (ws)% ⊦ str ∈ sdt[str, tree]; (s)