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Proof step 27 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), ⌜n⌝ < ⌜n0⌝ + # σ0, ⌜n⌝ < #((σ to m + n0) from m) + # σ0 ⊦ ((σ to n0 + m) + σ0)[m + n] = ((σ to m + n0) from m)[n], # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝ , ⌜m + n⌝ < #(σ to n0 + m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
#((σ to n0 + m) + σ0 from m) = #((σ to m + n0) from m) + # σ0, ⌜n⌝ < #((σ to m + n0) from m) + # σ0, ⌜n⌝ < ⌜n0⌝ + # σ0, ⌜n⌝ < #((σ to m + n0) from m), n0 ≤ n ⊦ σ0[n - n0] = ((σ to m + n0) from m)[n], # σ < ⌜m⌝ + ⌜n0⌝ , # σ < ⌜n0⌝ + ⌜m⌝

Rule argument

Additional information

Used simplifier rules: get-append-notls: ⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)[m] = σ1[m - (# σ).n];%size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ ⌜n⌝ + n∞1 < ⌜n⌝ + n∞2 ↔ n∞1 < n∞2; (s)% ⊦ m + n - n + n1 = m - n1; (s)%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf% ⊦ ⌜m⌝ .n = m; (s)%n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; (s)% ⊦ ⌜m⌝ ≠ ∞; (s)%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)