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Proof step 11 from proof stutf-alltaken (
previous step
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⊦ stutf(nσ) ↔ nσ ≠ [] ∧ nσ[0] = 0 ∧ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1)
Lemma
stutf-def