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Proof step 7 from proof mod-neq-03 (previous step)

m0 * n + n0 + 1 = m * n + n1, n1 + 1 < n, n0 < n, n1 < n0, n1 < n ⊦ n1 = 0, n0 + 1 < n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
(m0 * n0 + m0) + n0 + 1 = (m * n0 + m) + n1, n1 < n0 ⊦ n1 = 0

Rule argument

Additional information

Used simplifier rules: dist: ⊦ m0 * (m + n) = m0 * m + m0 * n;%one: ⊦ m0 * 1 = m0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;% ⊦ m < n → ¬ n < m; (s)% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat% ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)