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Proof step 3 from proof square-two (previous step)

m ≤ n ⊦ n * n + n = m * 2 ↔ n = m ∧ m < 2

User applied rule case distinction

And obtained the following new goals

Proof step 4:
n * n + n = m * 2, m ≤ n ⊦ n = m ∧ m < 2

Rule argument

pos: right 1

Additional information

Used simplifier rules: coef-17: ⊦ m * n + n = m1 * n ↔ m + 1 = m1 ∨ n = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≠ 1 → (m < 2 ↔ m = 0); (s)