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Proof step 2 from proof swap-to-from-app (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
n = n0 + m, m ≤ n ⊦ (σ to n) + σ0 from m = ((σ from m) to n0) + σ0, # σ < ⌜n⌝

Rule argument

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

Additional information

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