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Proof step 25 from proof max-zero-01 (previous step)

¬ ((1 + 1) * (1 + 2)) / 2 ≤ m, 1 ≤ m ⊦ 1 = 1 ↔ m = 1 ∨ m = 2, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
m < 3 ⊦ m = 0, m = 1, m = 2

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat