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Proof step 2 from proof sqle (previous step)

m * m ≤ m0 + 1, m0 * m0 < m ⊦ m0 = 0, m = 0

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: nat-mult inst: name: sqp seq: m0 ≠ 0, m0 ≠ 1, m0 * m0 ≤ m + 1 ⊦ m0 < m subst: [m0, m] → [m, m0] flags: current, []

Additional information

Used simplifier rules: zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0);%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%ls-08: ⊦ m * n + n0 < n ↔ m = 0 ∧ n0 < n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%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%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1 Used terms: m, m0