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Proof step 2 from proof square-inject (previous step)

m * (m + 1) + 2 * m0 = n * (n + 1) + 2 * n0, n0 ≤ n, m0 ≤ m ⊦ m = n ∧ m0 = n0

User applied rule apply lemma

And obtained the following new goals

Proof step 3:
m * (m + 1) ≤ n * (n + 1) + 2 * n0, n * (n + 1) + 2 * n0 < (m + 1) * (m + 2), n0 ≤ n ⊦ m = n
Proof step 4:
m * (m + 1) + 2 * m0 = n * (n + 1) + 2 * n0, m0 ≤ m, n0 ≤ n ⊦ m = n ∧ m0 = n0, m * (m + 1) ≤ n * (n + 1) + 2 * n0 ∧ n * (n + 1) + 2 * n0 < (m + 1) * (m + 2)

Rule argument

spec: <toplevel> name: square-interval seq: m * (m + 1) ≤ n * (n + 1) + 2 * n0, n * (n + 1) + 2 * n0 < (m + 1) * (m + 2), n0 ≤ n ⊦ m = n subst: [m, n, n0] → [m, n, n0]

Additional information

Used simplifier rules: dist: ⊦ m0 * (m + n) = m0 * m + m0 * n;%one: ⊦ m0 * 1 = m0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%zero-02: ⊦ m * n = m * n0 ↔ m = 0 ∨ n = n0;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s) Used terms: m, n, n0