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Proof step 3 from proof ls-07 (previous step)

n0 < n ⊦ n < m * n ↔ m ≠ 0 ∧ m ≠ 1

User applied rule constructor cut

And closed the goal!

Rule argument

var: m

Additional information

Used simplifier rules: zero: ⊦ 0 * n = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%dist: ⊦ m0 * (m + n) = m0 * m + m0 * n;%one: ⊦ m0 * 1 = m0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%zero: ⊦ 0 * n = 0;%zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0);%nat freely generated by 0, +1;;% ⊦ m +1 = m + 1; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m + 1 ≠ 0; (s)% ⊦ 1 = m + 1 ↔ m = 0; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ m < 0; (s)