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Proof step 7 from proof square-nle-01 (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 8:
n * n + n = m0 * 2 + n0, n0 < 2 ⊦ m0 ≤ n ↔ n < 2, m0 = 1, m0 = 0, n = 1, n = 0

Rule argument

pos: left 2

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)% ⊦ 2 * m = 0 ↔ m = 0; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)