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Proof step 59 from proof int2bytes-rec-neg-05 (previous step)

0 ≤ i * 8, # bits2bytes(restn((i→n(i * 8) + 1) % 8, int2bits(j))) = i→n(i), 0 ≤ 2 ^ (i * 8) + j, 0 < i, j + 2 ^ (i * 8 - 1) < 0, 0 < j + 2 ^ (i * 8), 0 < 2 ^ (i * 8) + j, j < 0 ⊦ bits2int(firstn((i→n(i * 8) + 1) % 8, int2bits(j))) = ~1, i2b(bits2int(firstn((i→n(i * 8) + 1) % 8, int2bits(j)))) = i2b(~1), (i→n(i * 8) + 1) % 8 = 0, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 60:
# bits2bytes(restn(i→n(i * 8) % 8, int2bits(j).rest)) = i→n(i), 0 ≤ 2 ^ (i * 8) + j, 0 < i, j + 2 ^ (i * 8 - 1) < 0, 0 < j + 2 ^ (i * 8), 0 < 2 ^ (i * 8) + j, j < 0 ⊦ (i→n(i * 8) + 1) % 8 = 0, j = 0, bits2int(true ' + firstn(i→n(i * 8) % 8, int2bits(j).rest)) = ~1, i2b(bits2int(true ' + firstn(i→n(i * 8) % 8, int2bits(j).rest))) = i2b(~1)

Rule argument

Additional information

Used simplifier rules: ⊦ i2b(i) ≠ i2b(j) → i ≠ j; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ j < 0 → int2bits(j).first;% ⊦ firstn(n + 1, bits) = bits.first + firstn(n, bits.rest);% ⊦ restn(n + 1, bits) = restn(n, bits.rest);%n ≠ 0 ⊦ (m + 1) % n ≠ 0 → (m + 1) % n = m % n + 1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;