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Proof step 4 from proof aligndown-sub-add-sub (previous step)

n1 ≤ n0 ⊦ n0 - n1 + ((n0 - n1) % n = 0 ⊃ n0 - n1;(n0 - n1) / n * n) = (n0 - n1) % n, n = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
n0 = n2 + n1, n1 ≤ n0 ⊦ n0 - n1 + (n2 % n = 0 ⊃ n2;n2 / n * n) = n2 % n, n = 0

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n0, n1, n2]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n0, n1, n2