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Proof step 7 from proof rounded-help2 (previous step)

n0 < m2, 2 * m2 ≤ n0 + m1, ∃ m0. m1 + m0 = m2 ∧ m0 ≠ 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n0 < m1 + m, 2 * (m1 + m) ≤ n0 + m1 ⊦ m = 0

Rule argument

Additional information

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