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Proof step 5 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 = m0 -1 ∧ n0 = n1 ⊦ n = 0, m1 + 1 = m0 ∧ n0 = n1

User applied rule case distinction

And obtained the following new goals

Proof step 6:
(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

Rule argument

pos: left 4

Additional information

Used simplifier rules: coef-17: ⊦ m * n + n = m1 * n ↔ ¬ ¬(m + 1 = m1 ∨ n = 0);%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%pred-03: ⊦ n -1 = n - 1; from specification nat%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2