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Proof step 26 from proof int2bytes-rec-neg-05 (previous step)

# int2bits(j) % 8 = 0 ∧ int2bits(j) ≠ [], # int2bits(j) = n * 8, # int2bits(j) / 8 = i→n(i) + 1, 0 < i, j + 2 ^ (i * 8 - 1) < 0, 0 < j + 2 ^ (i * 8), 0 < 2 ^ (i * 8) + j, j < 0 ⊦ j = 0, bits2bytes(int2bits(j)) = i2b(~1) ' + bits2bytes(int2bits(j)).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
# int2bits(j) = 8 * i→n(i) + 8, 0 < i, j + 2 ^ (i * 8 - 1) < 0, 0 < j + 2 ^ (i * 8), 0 < 2 ^ (i * 8) + j, j < 0 ⊦ j = 0, bits2bytes(int2bits(j)) = i2b(~1) ' + bits2bytes(int2bits(j)).rest

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m * (n + 1) = m * n + m;%n ≠ 0 ⊦ (m * n) % n = 0;% ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;%n ≠ 0 ⊦ (m * n) / n = m;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ int2bits(j) ≠ [];