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Proof step 2 from proof mod-neq (previous step)

⊦ m % n < (m + 1) % n ↔ (m + 1) % n ≠ 0, n = 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: mod-succ-02: n ≠ 0 ⊦ (m + 1) % n = 0 → m % n = n - 1;%mod-ls: n ≠ 0 ⊦ m % n < n;%mod-succ-06: n ≠ 0 ⊦ (m + 1) % n ≠ 0 → (m + 1) % n = m % n + 1;% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic