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Proof step 1 from proof int2bytes-rec-neg-06 (conclusion)

j < ~(2 ^ (i * 8 - 1)), 0 < i ⊦ int2bytes(j) = int2bytes((j - (2 ^ (i * 8) - 1)) / 2 ^ (i * 8)) + fix(i→n(i + 1), int2bytes(j % 2 ^ (i * 8))).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 < i, j + 2 ^ (i * 8 - 1) < 0 ⊦ int2bytes(j) = int2bytes((j + 1 - 2 ^ (i * 8)) / 2 ^ (i * 8)) + fix(i→n(i) + 1, int2bytes(j % 2 ^ (i * 8))).rest

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < ~ j ↔ i + j < 0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 < i → i→n(i + 1) = i→n(i) + 1;% ⊦ i - (j - j0) = i + j0 - j;