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Proof step 12 from proof tobuf-rev (previous step)

⌜n⌝ < # σ, (# σ).n ≤ n ⊦ # σ = ∞, ⌜σ.buf⌝ [n] = σ[n]

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat-inf name: elim seq: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ) subst: [n∞, m] → [# σ, m]

Additional information

Used simplifier rules: ⊦ ⌜m⌝ ≠ ∞; (s)% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: # σ, m