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Proof step 12 from proof le-same (previous step)

m < m * n + n0, n = 0 ∧ n0 ≤ m ∨ m = 0 ∧ n0 = 0 ∨ n = 1 ∧ n0 = 0 ⊦ m * n + n0 = 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 2

Additional information

Used simplifier rules: zero: ⊦ 0 * n = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%zero: ⊦ 0 * n = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%one: ⊦ m0 * 1 = m0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat