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Proof step 7 from proof le-succ (previous step)

m + 1 < m * n, m = 1 ∧ n ≤ 2 ∨ 2 ≤ m ∧ n = 1 ⊦ m = 0, n = 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 2

Additional information

Used simplifier rules: one: ⊦ m0 * 1 = m0;%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;%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat