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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 28:
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⌝

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, n0, n1] flags:

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, n0, n1