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Proof step 23 from proof int2bytes-rec-neg-05 (previous step)

8 ≠ 0, ∃ n0. # int2bits(j) = n0 * 8, # bits2bytes(int2bits(j)) = 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 24:
# int2bits(j) = n * 8, # bits2bytes(int2bits(j)) = 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

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;