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Proof step 2 from proof modeq (previous step)

n0 < n, n1 < n ⊦ m * n + n0 = m0 * n + n1 ↔ m = m0 ∧ n0 = n1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m * n + n0 = m0 * n + n1, n0 < n, n1 < n ⊦ m = m0 ∧ n0 = n1

Rule argument

Additional information

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