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Proof step 1 from proof disjoint-nojunk-newref-subset-01 (previous step)

⊦ disjoint(str, refs, st) ↔ ¬ (∃ refs1, refs2. refs1 ≠ [] ∧ refs2 ≠ [] ∧ refs1 .first ∈ refs ∧ refs2 .first ∈ refs ∧ refs1 .first ≠ refs2 .first ∧ refs1 .last = refs2 .last ∧ refs1 .last ≠ jvmref ∧ validrefs(str, refs1, st[refs1 .first -- _type] .type, st) ∧ validrefs(str, refs2, st[refs2 .first -- _type] .type, st))

Lemma disjoint-def