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Proof step 4 from proof cong-equiv (previous step)

¬ n ≤ n0 ⊦ m | dist(n, n0) ↔ n % m = n0 % m, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n0 < n ⊦ m = 0, m | dist(n, n0) ↔ n % m = n0 % m

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat