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Proof step 8 from proof to-unwind-end (previous step)

n = m + 1 ⊦ n = 0, ⌜a '⌝ + σ0 = [], σ0 to n = (σ0 to m) + ⌜(⌜a '⌝ + σ0)[n] '⌝ , # σ0 < ⌜n⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⌜m⌝ < # σ0 ⊦ ⌜a '⌝ + σ0 = [], σ0 to m + 1 = (σ0 to m) + ⌜σ0[m] '⌝

Rule argument

Additional information

Used simplifier rules: get-cons-succ: ⊦ (⌜a '⌝ + σ)[n + 1] = σ[n];%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%size-succ: ⊦ # σ < ⌜n⌝ + ⌜1⌝ ↔ ¬ ⌜n⌝ < # σ;%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2