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Proof step 2 from proof disjoint-nopath-both (
previous step
)
⊦ disjoint(str, r ' + r0 ', st) ∧ path∃(str, r, r1, st) ∧ path∃(str, r0, r2, st) ∧ r ≠ r0 → disjoint(str, r1 ' + r2 ', st)
Lemma
disjoint-pathex-forward-both