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Proof step 16 from proof abs-paths-eq (
previous step
)
⊦ ∃ p, p0, p1. p2 = p + p0 ∧ p3 = p + p1 ∧ (p0 ≠ ε ∧ p1 ≠ ε → p0.head ≠ p1.head)
A spec-lemma