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Proof step 29 from proof divi-same (previous step)

m * n2 + n < n0, m * n1 = n0 - m * n2 + n, m * n2 + n < n0 ⊦ n0 = 0, m | dist(n, 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] → [n0, m * n2 + n, n3] flags:

Additional information

Used simplifier rules: dist-add: ⊦ dist(m + n, m) = n;%dist-c: ⊦ dist(m, n) = dist(n, m);%divi-add-01: ⊦ m | m * n + n0 ↔ m | n0;%divi-mult: ⊦ m | m * n0;%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%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%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%zero-01: ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%lels-03: ⊦ m < n → m ≤ n; from specification nat%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n0, m * n2 + n, n3