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Proof step 19 from proof swap-to-from-app (previous step)

#((σ to n0 + m) + σ0 from m) = #((σ to m + n0) from m) + # σ0, ⌜n⌝ < #((σ to m + n0) from m) + # σ0, ⌜n0⌝ + # σ0 < #((σ to m + n0) from m) + # σ0, n0 ≤ n ⊦ ((σ to n0 + m) + σ0 from m)[n] = σ0[n - (#((σ to m + n0) from m)).n], # σ < ⌜m⌝ + ⌜n0⌝ , # σ < ⌜n0⌝ + ⌜m⌝ , ⌜n⌝ < ⌜n0⌝ + # σ0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: fin-size-from seq: ⊦ ⌜n⌝ < # σ ∧ finp(σ) → #(σ from n) = ⌜(# σ).n - n⌝ subst: [n, σ] → [m, σ to m + n0] flags: <all paths>

Additional information

Used simplifier rules: size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%fin-to: ⊦ finp(σ to m);%size-empty-equiv: ⊦ # σ = ⌜0⌝ ↔ σ = [];%ton-fromn: ⊦ ¬ # σ < ⌜n⌝ → (σ to n) from n = [];%size-empty: ⊦ # [] = ⌜0⌝ ;%size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%fin-to: ⊦ finp(σ to m);% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ ⌜m⌝ ≠ ∞; (s)%n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; (s)% ⊦ ⌜m⌝ .n = m; (s)%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% ⊦ m + n - m = n; (s)% ⊦ ⌜n⌝ < ⌜m⌝ + ⌜n0⌝ ↔ n < m + n0; (s)% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ n + 0 = n; (s)% ⊦ n∞ + ⌜0⌝ = n∞; (s)% ⊦ n∞1 < n∞ + n∞1 ↔ n∞ ≠ ⌜0⌝ ∧ n∞1 ≠ ∞; (s)% ⊦ 0 ≤ m; (s)% ⊦ m - 0 = m; (s) Used terms: m, σ to m + n0