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Proof step 2 from proof modls-01 (previous step)

n2 < n0, m1 < m, n1 < n, n ≤ n0 ⊦ m1 * n + n1 < m * n0 + n2

User applied rule apply lemma

And obtained the following new goals

Proof step 3:
⊦ m ≤ n ∧ n < n0 → m < n0

Rule argument

spec: nat name: fle-01 seq: ⊦ m ≤ n ∧ n < n0 → m < n0 subst: [n, m, n0] → [m1 * n0 + n1, m1 * n + n1, m * n0 + n2]

Additional information

Used simplifier rules: mod-nle: ⊦ m0 < m ∧ n ≤ n0 ∧ n2 < n0 ∧ n1 < n → ¬ m * n0 + n2 ≤ m0 * n + n1;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;% ⊦ m < n ↔ ¬ n ≤ m; (ss)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m1 * n0 + n1, m1 * n + n1, m * n0 + n2