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Proof step 126 from proof refine-readmeta (previous step)

∀ p. p ∈ ps → fpathtolink(p) ∈ ls, ∀ alink. alink ∈ ls → (∃ p. p ∈ ps ∧ fpathtolink(p) = alink), ∀ p, q. p ∈ ps ∧ q ∈ ps ∧ fpathtolink(p) = fpathtolink(q) → p = q ⊦ # ps = # ls ↔ true

A spec-lemma