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Proof step 30 from proof swap-to-from-app (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
#((σ to n0 + m) + σ0 from m) = #((σ to m + n0) from m) + # σ0, ⌜n⌝ < #((σ to m + n0) from m) + # σ0, n < n0 ⊦ σ[m + n] = ((σ to m + n0) from m)[n], n0 = 0, # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

Rule argument

Additional information

Used simplifier rules: size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%get-to: m < n, ¬ # σ < ⌜n⌝ ⊦ (σ to n)[m] = σ[m];%less-size-from: ⌜n⌝ < # σ ⊦ ⌜n0⌝ < #(σ from n) ↔ ⌜n0 + n⌝ < # σ;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%ls-lsadd: ⊦ m < n → ⌜m⌝ < ⌜n⌝ + n∞0; from specification nat-inf%zero-01: ⊦ ¬ m < 0; from specification nat-basic1%add-all-zero: ⊦ n∞ + ⌜0⌝ = n∞; from specification nat-inf%add-base: ⊦ n + 0 = n; from specification nat-basic2%le-add: ⊦ m + m0 ≤ m ↔ m0 = 0; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%ls-natinf-to-nat-1-2: ⊦ ⌜n⌝ < ⌜m⌝ + ⌜n0⌝ ↔ n < m + n0; from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%ls-natinf-to-nat-2-2: ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ + ⌜n1⌝ ↔ n + m < n0 + n1; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf