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Proof step 68 from proof at-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]) → dnode(md2, sdt2[str2, tree])[q] = (str2 ' ⊑ q ⊃ tree[q.tail];(q = ε ⊃ dnode(md1, sdt1[str2, tree]);dnode(md1, sdt1)[q])) ⊦ dnode(md, sdt[str, tree0])[q0] = dnode(md, sdt)[q0], q0 = ε, 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