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Proof step 4 from proof divi-ls (previous step)

m < n0 ⊦ (∃ n1. n0 * n1 = m) ↔ m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m < n0 ⊦ (∃ n1. n0 * n1 = 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