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Proof step 5 from proof mod-nls (previous step)

n1 < n, n2 < n0, m0 < m, m * n0 + n2 < m0 * n + n1, n1 < n0, n ≤ n0, m0 < m ∧ n ≤ n0 ∧ n2 < n0 ∧ n1 < n → ¬ m * n0 + n2 ≤ m0 * n + n1 ⊦

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 7

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)%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% ⊦ m +1 = m + 1; (s)% ⊦ ¬ n < n; (s)