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Proof step 2 from proof ls-07 (previous step)

n0 < n ⊦ n < m * n ↔ 1 < m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n0 < n ⊦ n < m * n ↔ m ≠ 0 ∧ m ≠ 1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)