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Proof step 3 from proof to-unwind-end (previous step)

σ ≠ [] → (∃ a, σ0. σ = ⌜a '⌝ + σ0), ⌜n⌝ < # σ ⊦ σ to n + 1 = (σ to n) + ⌜σ[n] '⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ ⌜a '⌝ + σ0 = [], ⌜a '⌝ + (σ0 to n) = (⌜a '⌝ + σ0 to n) + ⌜(⌜a '⌝ + σ0)[n] '⌝ , # σ0 < ⌜n⌝

Rule argument

Additional information

Used simplifier rules: size-empty: ⊦ # [] = ⌜0⌝ ;%to-succ-cons: ⊦ ⌜a '⌝ + σ to n + 1 = a + (σ to n);%cons-append: ⊦ a + σ = ⌜a '⌝ + σ;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%size-append: ⊦ #(σ + σ1) = # σ + # σ1;%size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ;%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%less-succ-leq: ⊦ ⌜n⌝ < ⌜1⌝ + n∞ ↔ ¬ n∞ < ⌜n⌝ ; from specification nat-inf%len-02: ⊦ # a ' = 0 +1; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%zero-01: ⊦ ¬ m < 0; from specification nat-basic1%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf