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Proof step 38 from proof int2bytes-rec-neg-05 (previous step)

# int2bits(j) % 8 = 0 ∧ 8 < # int2bits(j), # 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(firstn(8, int2bits(j))) + bits2bytes(restn(8, int2bits(j))) = i2b(~1) ' + (bits2bytes(firstn(8, int2bits(j))) + bits2bytes(restn(8, int2bits(j)))).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
# int2bits(j) = 8 * i→n(i) + 8, # int2bits(j) % 8 = 0, j < 0, 0 < 2 ^ (i * 8) + j, 0 < j + 2 ^ (i * 8), j + 2 ^ (i * 8 - 1) < 0, 0 < i, 8 < # int2bits(j) ⊦ bits2bytes(firstn(8, int2bits(j))) + bits2bytes(restn(8, int2bits(j))) = i2b(~1) ' + (bits2bytes(firstn(8, int2bits(j))) + bits2bytes(restn(8, int2bits(j)))).rest, j = 0, i = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0;% ⊦ m < m + n ↔ n ≠ 0;% ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;