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Proof step 40 from proof in-put (previous step)

sdt0[str0] = dnode(md, sdt), str0 ∈ sdt0, q0 ∈ dnode(md, sdt[str, tree0]), ∀ 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 ' ⊑ q0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: md, sdt, md, sdt, q0, str, tree0

Additional information

Used terms: md, sdt, md, sdt, q0, str, tree0