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Proof step 7 from proof square-two (previous step)

n * n + n = m * 2, m ≤ n, n = 1 ∨ n = 0 ⊦ n = m ∧ m < 2

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 3

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%twoeq: ⊦ 2 = 2 * n ↔ n = 1;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%two-zero-01: ⊦ 2 * m = 0 ↔ m = 0;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat