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Proof step 6 from proof dnode-onein (previous step)

str ∈ sdt0 ⊦ ∃ md, sdt. dnode(md0, sdt0) = dnode(md, sdt) ∧ str ∈ sdt

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: md0, sdt0

Additional information

Used terms: md0, sdt0