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Proof step 4 from proof is-aligned-minus (previous step)

n1 % m = 0, n0 % m = 0, n1 ≤ n0 ⊦ (n0 - n1) % m = 0, m = 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: ⊦ m0 % n = 0 ∧ 0 ≠ n → (m + m0) % n = m % n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n0, n1, n