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Proof step 10 from proof plinksd+ (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 11:
p ∈ links(fid, dnode(md, sdt[str, tree])) ⊦ str ∈ sdt, p ∈ links(fid, dnode(md, sdt)), p ≠ ε ∧ p.head = str ∧ p.tail ∈ links(fid, tree)
Proof step 21:
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]))

Rule argument

pos: right 1

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