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Proof step 5 from proof cong-equiv (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 < n ⊦ m | dist(n, n0) ↔ n % m = n0 % m, m = 0

Rule argument

Additional information

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