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Proof step 18 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 < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝ , ⌜n⌝ < ⌜n0⌝ + # σ0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

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);%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic%prime-add-fin: n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; from specification nat-inf%sel: ⊦ ⌜m⌝ .n = m; from specification nat-inf-basic%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%eq-01: ⊦ m + n - m = n; from specification nat%ls-natinf-to-nat-1-2: ⊦ ⌜n⌝ < ⌜m⌝ + ⌜n0⌝ ↔ n < m + n0; from specification nat-inf%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2%lez: ⊦ 0 ≤ m; from specification nat%add-all-zero: ⊦ n∞ + ⌜0⌝ = n∞; from specification nat-inf%add-less: ⊦ n∞1 < n∞ + n∞1 ↔ ¬ ¬(n∞ ≠ ⌜0⌝ ∧ n∞1 ≠ ∞); from specification nat-inf%sub-base: ⊦ m - 0 = m; from specification nat Used terms: m, σ to m + n0