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Proof step 31 from proof cong-equiv (previous step)

n = m0 * m + n0 ∧ n0 < m, n0 = (n1 + n) % m ⊦ m | n1, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
n0 = (n1 + n0) % m, n0 < m ⊦ m | n1

Rule argument

Additional information

Used simplifier rules: mod-ls: n ≠ 0 ⊦ m % n < n; from specification nat-div%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mod-mult: n ≠ 0 ⊦ (m + n0 * n) % n = m % n; from specification nat-div