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Proof step 6 from proof square-nle-01 (previous step)

n * n + n = m0 * 2 + n0 ∧ n0 < 2 ∧ 2 ≠ 0, m0 ≠ 0 ∧ m0 ≠ 1 ↔ n ≠ 0 ∧ n ≠ 1 ⊦ m0 ≤ n ↔ n < 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n * n + n = m0 * 2 + n0, m0 ≠ 0 ∧ m0 ≠ 1 ↔ n ≠ 0 ∧ n ≠ 1, n0 < 2 ⊦ m0 ≤ n ↔ n < 2

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1