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Proof step 9 from proof enc-invert (previous step)

(maxenc(m) * (maxenc(m) + 1)) % 2 = 0 ∧ 2 ≠ 0 → (∃ n0. maxenc(m) * (maxenc(m) + 1) = 2 * n0) ⊦ ( ((maxenc(m) - (m - (maxenc(m) * (maxenc(m) + 1)) / 2)) + (m - (maxenc(m) * (maxenc(m) + 1)) / 2)) * ((maxenc(m) - (m - (maxenc(m) * (maxenc(m) + 1)) / 2)) + (m - (maxenc(m) * (maxenc(m) + 1)) / 2) + 1)) / 2 + (m - (maxenc(m) * (maxenc(m) + 1)) / 2) = m

User applied rule case distinction

And obtained the following new goals

Proof step 10:
∃ n0. maxenc(m) * (maxenc(m) + 1) = 2 * n0 ⊦ ( ((maxenc(m) - (m - (maxenc(m) * (maxenc(m) + 1)) / 2)) + (m - (maxenc(m) * (maxenc(m) + 1)) / 2)) * ((maxenc(m) - (m - (maxenc(m) * (maxenc(m) + 1)) / 2)) + (m - (maxenc(m) * (maxenc(m) + 1)) / 2) + 1)) / 2 + (m - (maxenc(m) * (maxenc(m) + 1)) / 2) = m

Rule argument

pos: left 1

Additional information

Used simplifier rules: max-nls: ⊦ ¬ m < (maxenc(m) * maxenc(m) + maxenc(m)) / 2;% ⊦ m * (n + 1) = m * n + m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m ≤ n ↔ ¬ n < m; (ss)%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ (m * m + m) % 2 = 0; (s)