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Proof step 28 from proof swap-to-from-app (previous step)

n = n1 + n0, #((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, n0 ≤ n ⊦ σ0[n1] = ((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 29:
#((σ to n0 + m) + σ0 from m) = #((σ to m + n0) from m) + # σ0, ⌜n1⌝ + ⌜n0⌝ < #((σ to m + n0) from m), ⌜n1⌝ + ⌜n0⌝ < #((σ to m + n0) from m) + # σ0, ⌜n1⌝ < # σ0 ⊦ σ0[n1] = ((σ to m + n0) from m)[n1 + n0], # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

Rule argument

Additional information

Used simplifier rules: lep: ⊦ m ≤ m + n; from specification nat%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%add-ls-cancel-01: ⊦ ⌜n⌝ + n∞1 < ⌜n⌝ + n∞2 ↔ n∞1 < n∞2; 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