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Proof step 5 from proof mult-01 (previous step)

n0 + 1 < n, ∀ m0. m0 | m * n → m0 = 1 ∨ m0 = m * n ⊦ m = 1, n = 0, m * n = 1, m = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: m

Additional information

Used simplifier rules: divi-mult: ⊦ m | m * n0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%eq-same: ⊦ m * n = m ↔ ¬ ¬(m = 0 ∨ n = 1); from specification nat-mult%one-01: ⊦ m0 * m = 1 ↔ m0 = 1 ∧ m = 1; from specification nat-mult%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%l-05: ⊦ m + m0 < n → ¬ n + n0 < m; from specification nat-basic2 Used terms: m