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Proof step 2 from proof less-size-from (previous step)

⌜n⌝ < # σ ⊦ ⌜n0⌝ < #(σ from n) ↔ ⌜n0⌝ + ⌜n⌝ < # σ

User applied rule structural induction

And closed the goal!

Rule argument

indvar: σ

Additional information

Used simplifier rules: from-fin: ⊦ ⌜x⌝ from n = ⌜x from n⌝ ;%size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ;%length-from: ¬ # x < n ⊦ #(x from n) = # x - n;%from-inf: ⊦ ⌜nf⌝ from n = ⌜nf from n⌝ ;%size-inf: ⊦ # ⌜nf⌝ = ∞;%stream freely generated by ⌜ ⌝, ⌜ ⌝;;%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf%inf-less: ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; from specification nat-inf%addeqinf-cancel: ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; from specification nat-inf%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%ls: n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n; from specification nat%ls-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2