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Proof step 35 from proof int2bytes-first-neg (previous step)

¬(i + 1 < 0 ∧ j < 255 ∧ 0 < 255), int2bytes(i) = by ' + bytes0, i ≤ 0, j ≤ 0, 0 ≤ b2i(by), i * 256 + j < ~383, abs(j) < 256, by ∈ int2bytes(i), 0 ≤ (i + 1) * 255 + j, i * 256 + j < 0, i * 256 < (i + 1) * 255, ∀ bytes, by, i0. i→n(i0) < i→n((i * 256 + j) + 255) → int2bytes(i0) = by ' + bytes ∧ by ∈ int2bytes(i0) ∧ 0 ≤ b2i(by) ∧ i0 < 0 → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
int2bytes(i) = by ' + bytes0, i ≤ 0, j ≤ 0, 0 ≤ b2i(by), i * 256 + j < ~383, abs(j) < 256, by ∈ int2bytes(i), 0 ≤ (i + 1) * 255 + j, i * 256 + j < 0, i * 256 < (i + 1) * 255, ~1 ≤ i, ∀ bytes, by, i0. i→n(i0) < i→n((i * 256 + j) + 255) → int2bytes(i0) = by ' + bytes ∧ by ∈ int2bytes(i0) ∧ 0 ≤ b2i(by) ∧ i0 < 0 → false ⊦

Rule argument

Additional information

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