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Proof step 3 from proof cyclic-subset (
previous step
)
⊦ validrefs(str, refs, ty, st0) ∧ st ⊆ st0 ∧ ¬ junkref(str, r, st) ∧ refs ≠ [] ∧ refs .first = r → validrefs(str, refs, ty, st)
Lemma
validrefs-subset