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Proof step 6 from proof dist-mult (previous step)

n0 < n, n0 < n ⊦ dist(m * n, m * n0) = m * (n - n0), n = 0

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] → [n, n0, n1] flags:

Additional information

Used simplifier rules: dist-mult-01: ⊦ dist(m * (n1 + n0), m * n0) = m * n1;%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%zero-02: ⊦ m * n = m * n0 ↔ m = 0 ∨ n = n0; from specification nat-mult%lels-03: ⊦ m < n → m ≤ n; from specification nat%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, n0, n1