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Proof step 6 from proof mult-le-01 (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 3

Additional information

Used simplifier rules: sqr-08: ⊦ m * m ≤ 2 ↔ m ≤ 1;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) 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%twols-02: ⊦ m ≠ 0 → (m < 2 ↔ m = 1); from specification nat-basic1%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat