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Proof step 5 from proof div-succ (previous step)

n0 + 1 = n, m = m0 * n + n0, n0 < n ⊦ (m + 1) / n = m0 + 1, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ (m0 * n0 + m0) / (n0 + 1) = m0

Rule argument

Additional information

Used simplifier rules: add: n ≠ 0 ⊦ (n0 + n) / n = n0 / n + 1;% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m * (n + 1) = m * n + m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)