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Proof step 35 from proof swap-to-from-app (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 36:
#((σ to n0 + m) + σ0 from m) = #((σ to m + n0) from m) + # σ0, #((σ to m + n0) from m) = ⌜n1⌝ , #((σ to m + n0) from m) < ⌜n0⌝ + # σ0, ⌜n⌝ < #((σ to m + n0) from m) + # σ0, ⌜n⌝ < ⌜n0⌝ + # σ0 ⊦ #((σ to m + n0) from m) = ∞, ((σ to n0 + m) + σ0)[m + n] = σ0[n - n1], σ0 = [], ⌜n⌝ < #((σ to m + n0) from m), # σ < ⌜m⌝ + ⌜n0⌝ , # σ < ⌜n0⌝ + ⌜m⌝

Rule argument

spec: nat-inf name: elim seq: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ) subst: [n∞, m] → [#((σ to m + n0) from m), m0]

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: #((σ to m + n0) from m), m0