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Proof step 2 from proof two-03 (previous step)

n ≤ 2 ⊦ even(n) ↔ n = 2, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
2 * m ≤ 2 ⊦ m = 1, m = 0

Rule argument

Additional information

Used simplifier rules: Even: ⊦ even(n) ↔ (∃ m. m + m = n);% ⊦ 2 * m = 0 ↔ m = 0; (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ 2 = 2 * n ↔ n = 1; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + m = 2 * m; (s)