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Proof step 6 from proof leone (previous step)

n0 * m + m ≤ n0 + 1, ∀ m. m * n0 ≤ n0 ↔ m ≤ 1 ⊦ n0 = 0, m = 0, m = 1

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: m abort: F discard: T computed: F

Additional information

Used simplifier rules: zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0);%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%coef-18: ⊦ m * n + n ≤ m1 * n ↔ ¬ ¬(m + 1 ≤ m1 ∨ n = 0);%one-02: ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1); from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%ref-1: ⊦ ¬ n < n; from specification nat-basic1 Used terms: m