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Proof step 4 from proof ls-01 (previous step)

m1 * m0 + m0 ≤ m1, ∀ m0. m0 * m1 < m1 → m1 = 0 ∨ m0 = 0 ⊦ m0 = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: m0

Additional information

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