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Proof step 3 from proof sqle (previous step)

m0 * m0 < m, m * m ≤ m0 + 1 ⊦ m0 = 0

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
m0 ≠ 0, m0 ≠ 1, m0 * m0 ≤ m + 1 ⊦ m0 < m

Rule argument

spec: <toplevel> name: sqp seq: m0 ≠ 0, m0 ≠ 1, m0 * m0 ≤ m + 1 ⊦ m0 < m subst: [m0, m] → [m, m0]

Additional information

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