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Proof step 9 from proof mod-succ-04 (previous step)

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

Heuristic pl case distinction applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ 0 < m ↔ m ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ m ≠ 1 → m * n0 ≠ m * m0 + 1; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic% ⊦ m +1 = m + 1; (s)% ⊦ n1 < n0 ∧ n0 < n ∧ n1 ≠ 0 → m0 * n + n0 + 1 ≠ m * n + n1; (s)% ⊦ m0 * 1 = m0; (s)% ⊦ ¬ m < n → ¬ m + m0 < n; (s)