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Proof step 6 from proof modeq (previous step)

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

User applied rule case distinction

And obtained the following new goals

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

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;% ⊦ n -1 = n - 1; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ m < n → n ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m + 1 ≠ 0; (s)%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s)