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Proof step 6 from proof modeq (previous step)

(m1 * n + n) + n0 = m0 * n + n1, n0 < n, n1 < n ⊦ m1 * n + n0 = m0 -1 * n + n1 ∧ n0 < n ∧ n1 < n, n = 0, m1 + 1 = m0 ∧ n0 = n1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
(m1 * n + n) + n0 = m0 * n + n1, n1 < n, n0 < n ⊦ m1 * n + n0 = (m0 - 1) * n + n1, n = 0, m1 + 1 = m0

Rule argument

Additional information

Used simplifier rules: rsucc-01: ⊦ (m + 1) * n = m * n + n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%add-base: ⊦ n + 0 = n; from specification nat-basic2%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%pred-03: ⊦ n -1 = n - 1; from specification nat