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Proof step 5 from proof le-04 (previous step)

m ≤ n ⊦ n * n0 ≤ m ↔ n0 = 1 ∧ m = n ∨ n = 0, n0 = 0

User applied rule cut formula

And closed the goal!

Rule argument

fma: n0 = 1

Additional information

Used simplifier rules: one: ⊦ m0 * 1 = m0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%mult-base: ⊦ m * 0 = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%leone: ⊦ n ≠ 0 → (m * n ≤ n ↔ m ≤ 1);% ⊦ 0 ≤ m; (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1); (s)% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ m ≤ 0 ↔ m = 0; (s) Used formulas: n0 = 1