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Proof step 3 from proof append-from-ge-02 (previous step)

¬ # σ0 < ⌜m⌝ ⊦ ¬ ⌜m⌝ < # σ0 → σ0 + σ from m = σ from (m - (# σ0).n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⌜m⌝ = # σ0 ⊦ σ0 + σ from m = σ

Rule argument

Additional information

Used simplifier rules: from-zero: ⊦ σ from 0 = σ;%ler: ⊦ m ≤ m; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%zero: ⊦ m - m = 0; from specification nat%sel: ⊦ ⌜m⌝ .n = m; from specification nat-inf-basic%lele-refl: ⊦ ¬ n∞1 < n∞ → (¬ n∞ < n∞1 ↔ n∞ = n∞1); from specification nat-inf