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Proof step 3 from proof ls-01 (previous step)

m0 * m1 +1 < m1 +1, ∀ m0. m0 * m1 < m1 → m0 = 0 ⊦ m0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m0 * m1 + m0 ≤ m1, ∀ m0. m0 * m1 < m1 → m0 = 0 ⊦ m0 = 0

Rule argument

Additional information

Used simplifier rules: 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;% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)