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Proof step 21 from proof plinksd+ (previous step)

p ∈ links(fid, dnode(md, sdt)) ∨ p ≠ ε ∧ p.head = str ∧ p.tail ∈ links(fid, tree) ⊦ str ∈ sdt, p ∈ links(fid, dnode(md, sdt[str, tree]))

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information