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Proof step 2 from proof sqr-07 (previous step)

⊦ m * m ≤ m + 1 ↔ m ≤ 1

User applied rule constructor cut

And closed the goal!

Rule argument

var: m

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%rsucc-01: ⊦ (m + 1) * n = m * n + n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%two-02: ⊦ m + m = 2 * m;%numeral-20: m2 ≠ 0 ⊦ (* m2) * m + n ≤ m + n0 ↔ (m2 - 1) * m + n ≤ n0;%one: ⊦ m0 * 1 = m0;%nat freely generated by 0, +1;;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%le-num-01: n < n0 ⊦ m + (* n) ≤ m0 + (* n0) ↔ m ≤ m0 + (n0 - n); from specification nat%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%one-01: ⊦ m ≠ 0 → (m + n ≤ 1 ↔ m = 1 ∧ n = 0); from specification nat