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Proof step 7 from proof ls-03 (previous step)

n ≠ 0 → n0 / n * n ≤ n0 ∧ n0 < (n0 / n)+1 * n, n0 < n * (n0 + n) / n, (n0 + n) / n * n ≤ n0 + n, n0 / n + 1 < (n0 + n) / n ⊦ (n0 + n) / n = 0, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n0 < n * n0 / n + n, n0 / n * n ≤ n0, n0 / n + 1 < (n0 + n) / n, (n0 + n) / n * n ≤ n0 + n, n0 < n * (n0 + n) / n, n0 / n < (n0 + n) / n ⊦ n * n0 / n + n = 0, n = 0, (n0 + n) / n = 0

Rule argument

Additional information

Used simplifier rules: ls-02: ⊦ m0 ≠ 0 → (m0 * m < m0 * n ↔ m < n); from specification nat-mult%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%coef-21: ⊦ m * n < m1 * n + n ↔ ¬ ¬(m < m1 + 1 ∧ n ≠ 0); from specification nat-mult%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%rsucc-02: ⊦ m * (n + 1) = m * n + m; from specification nat-mult%succ: ⊦ m +1 = m + 1; from specification nat-basic2