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Proof step 4 from proof int2bytes-rec-neg-06 (previous step)

¬(j < ~(2 ^ (i * 8 - 1)) ∧ 0 < i ∧ j % 2 ^ (i * 8) ≠ 0), 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
j % 2 ^ (i * 8) = 0, j + 2 ^ (i * 8 - 1) < 0, 0 < i ⊦ int2bytes(j) = int2bytes((j + 1 - 2 ^ (i * 8)) / 2 ^ (i * 8)) + fix(i→n(i) + 1, i2b(0) ').rest

Rule argument

Additional information

Used simplifier rules: int2bytes-byte: ~128 ≤ i, i < 128 ⊦ int2bytes((* i)) = i2b(i) ';% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i < ~ j ↔ i + j < 0;