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Proof step 5 from proof mod-succ-01 (previous step)

m = m0 * n + n0, n0 < n ⊦ (m + 1) % n = n0 + 1, n0 + 1 = n, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 + 1 < n, n0 < n ⊦ ((m0 * n + n0) + 1) % n = n0 + 1, n0 + 1 = n

Rule argument

Additional information

Used simplifier rules: succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)