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Proof step 41 from proof swap-to-from-app (previous step)

#((σ to m + n0) from m) = ⌜n1⌝ , #((σ to n0 + m) + σ0 from m) = ⌜n1⌝ + # σ0, ⌜m + n2 + n1⌝ < #(σ to n0 + m), ⌜n2⌝ + ⌜n1⌝ < ⌜n0⌝ + # σ0, ⌜n2⌝ < # σ0, ⌜n1⌝ < ⌜n0⌝ + # σ0 ⊦ ((σ to n0 + m) + σ0)[m + n2 + n1] = σ0[n2], σ0 = [], # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
#((σ to n0 + m) + σ0 from m) = ⌜n1⌝ + # σ0, #((σ to m + n0) from m) = ⌜n1⌝ , ⌜n1⌝ < ⌜n0⌝ + # σ0, ⌜n2⌝ < # σ0, ⌜n2⌝ + ⌜n1⌝ < ⌜n0⌝ + # σ0, n1 + n2 < n0 ⊦ σ[m + n2 + n1] = σ0[n2], n0 = 0, σ0 = [], # σ < ⌜m⌝ + ⌜n0⌝ , # σ < ⌜n0⌝ + ⌜m⌝

Rule argument

Additional information

Used simplifier rules: get-append-ls: ⊦ ⌜m⌝ < # σ → (σ + σ1)[m] = σ[m];%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%get-to: m < n, ¬ # σ < ⌜n⌝ ⊦ (σ to n)[m] = σ[m];%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf% ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s)% ⊦ ⌜n⌝ + n∞1 < ⌜n⌝ + n∞2 ↔ n∞1 < n∞2; (s)%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic