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Proof step 229 from proof int2bytes-rec-neg-05 (previous step)

¬ ( #(mklist(true, 8 * i→n(i) + 8 - # int2bits(j % 2 ^ (i * 8))) + int2bits(j % 2 ^ (i * 8))) % 8 = 0 ∧ 8 < #(mklist(true, 8 * i→n(i) + 8 - # int2bits(j % 2 ^ (i * 8))) + int2bits(j % 2 ^ (i * 8)))), # int2bits(j % 2 ^ (i * 8)) < 8 * i→n(i), j + 2 ^ (i * 8 - 1) < 0, j + 2 ^ (i * 8) ≤ 0, j + 2 ^ (i * 8) < i, j % 2 ^ (i * 8) < 0, 0 < i, # int2bits(j % 2 ^ (i * 8)) ≤ 8 * i→n(i) + 8 ⊦ i = 0, 1 + 8 * i→n(i) = # int2bits(j % 2 ^ (i * 8)), mklist(true, 8 * i→n(i) - # int2bits(j % 2 ^ (i * 8))) + int2bits(j % 2 ^ (i * 8)) = [], j % 2 ^ (i * 8) = 0, bits2bytes(mklist(true, 8 * i→n(i) - # int2bits(j % 2 ^ (i * 8))) + int2bits(j % 2 ^ (i * 8))) = bits2bytes(mklist(true, 8 * i→n(i) + 8 - # int2bits(j % 2 ^ (i * 8))) + int2bits(j % 2 ^ (i * 8))).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m < n → n ≠ 0;% ⊦ m + n ≤ m ↔ n = 0;% ⊦ m + n ≤ m + n0 ↔ n ≤ n0;%n ≤ m ⊦ (m - n) + m0 ≤ n0 ↔ m + m0 ≤ n0 + n;% ⊦ m + n = n + m;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # mklist(boolvar, n) = n;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ m * n = n * m;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;