[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 12 from proof alignup-le-is-aligned (previous step)

m % n = 0, m < n * m0 / n + n, m0 < n * m0 / n + n, m0 ≤ m ⊦ m0 - m0 / n * n = 0, n = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
m % n = 0, m0 = n0 + m0 / n * n, m0 < n * m0 / n + n, m < n * m0 / n + n, m0 ≤ m, m0 / n * n ≤ m0 ⊦ n = 0, n0 = 0

Rule argument

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

Additional information

Used simplifier rules: n ≠ 0 ⊦ m / n * n ≤ m; (s)%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, m0 / n * n, n0