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Proof step 21 from proof divi-same (previous step)

n0 ≤ m0, m * n1 = m0 - n0 ⊦ m * n2 + m0 = 0, m = 0, n2 = 0, m | dist(m * n2 + m0, n0)

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m0, n0, n] flags:

Additional information

Used simplifier rules: dist-add: ⊦ dist(m + n, m) = n;%divi-add-01: ⊦ m | m * n + n0 ↔ m | n0;%divi-mult: ⊦ m | m * n0;%lep: ⊦ m ≤ m + n; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m0, n0, n