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Proof step 3 from proof modeq (previous step)

∀ n, n0, m0, n1. m1 * n + n0 = m0 * n + n1 ∧ n0 < n ∧ n1 < n → n = 0 ∨ m1 = m0 ∧ n0 = n1, m1 +1 * n + n0 = m0 * 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 4:
(m1 * n + n) + n0 = m0 * n + n1, n0 < n, n1 < n, ∀ n, n0, m0, n1. m1 * n + n0 = m0 * n + n1 ∧ n0 < n ∧ n1 < n → n = 0 ∨ m1 = m0 ∧ n0 = n1 ⊦ n = 0, m1 + 1 = m0 ∧ n0 = n1

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;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2