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Proof step 24 from proof swap-from-to (previous step)

#(σ from m) < ⌜n⌝ ⊦ #((σ from m) to n) = #((σ to m + n) from m), # σ < ⌜m⌝ + ⌜n⌝

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] flags: <all paths>

Additional information

Used simplifier rules: from-fin: ⊦ ⌜x⌝ from n = ⌜x from n⌝ ;%to-fin: ⊦ ⌜x⌝ to n = ⌜x to n⌝ ;%size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ;%length-from: ¬ # x < n ⊦ #(x from n) = # x - n;%length-to: ¬ # x < n ⊦ #(x to n) = n;% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%stream-fin-forward: ⊦ finp(σ) → (∃ x. ⌜x⌝ = σ); (forward) from specification stream-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ finp(⌜x⌝ ); (s)% ⊦ m < n → ¬ n < m; (s)% ⊦ ⌜m⌝ .n = m; (s)% ⊦ m < n → m ≤ n; (s)%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)% ⊦ ⌜n⌝ < ⌜m⌝ + ⌜n0⌝ ↔ n < m + n0; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (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%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n - m = n; (s)% ⊦ ⌜m⌝ = ⌜m0⌝ ↔ m = m0; (s) Used terms: m, σ