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Proof step 34 from proof int2bytes-first-neg-01 (previous step)

¬(255 * (i + 1) < 0 ∧ j < 0), int2bytes(i) = i2b(0) ' + bytes0, i * 256 < 255 * (i + 1), i * 256 + j < 0, 0 ≤ 255 * (i + 1) + j, i2b(0) ∈ int2bytes(i), abs(j) < 256, i * 256 + j < ~383, j ≤ 0, i ≤ 0, ∀ bytes, i0. i→n(i0) < i→n((i * 256 + j) + 255) → int2bytes(i0) = i2b(0) ' + bytes ∧ i0 < 0 ∧ i2b(0) ∈ int2bytes(i0) → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
int2bytes(i) = i2b(0) ' + bytes0, j < 0, ~256 < j, i2b(0) ∈ int2bytes(i), 0 ≤ 255 * (i + 1) + j, i * 256 + j < 0, i * 256 < 255 * (i + 1), ~1 ≤ i, i ≤ 0, i * 256 + j < ~383, ∀ bytes, i0. i→n(i0) < i→n((i * 256 + j) + 255) → int2bytes(i0) = i2b(0) ' + bytes ∧ i0 < 0 ∧ i2b(0) ∈ int2bytes(i0) → false ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ 0 < i ∧ j < 0 → ((* i) * i0 < (* j) ↔ i0 < j / i);% ⊦ i + 0 = i;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ 0 < i → (i * j < 0 ↔ j < 0);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;