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Proof step 16 from proof int2bytes-first-neg (previous step)

i = i1 + 255, int2bytes(i1 / 256) + int2bytes(i % 256).last ' = by ' + bytes, i < ~128, 0 ≤ b2i(by), i < 0, ∀ bytes, by, i0. i→n(i0) < i→n(i) → int2bytes(i0) = by ' + bytes ∧ by ∈ int2bytes(i0) ∧ 0 ≤ b2i(by) ∧ i0 < 0 → false ⊦ int2bytes(i % 256).last ≠ by ∧ ¬ int2bytes(i % 256).last ∈ bytes, ¬ by ∈ int2bytes(i1 / 256) ∧ by ≠ int2bytes(i % 256).last

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
int2bytes(i1 / 256) + int2bytes((i1 + 255) % 256).last ' = by ' + bytes, 0 ≤ b2i(by), i1 < ~383, i1 < 0, ∀ bytes, by, i0. i→n(i0) < i→n(i1 + 255) → int2bytes(i0) = by ' + bytes ∧ by ∈ int2bytes(i0) ∧ 0 ≤ b2i(by) ∧ i0 < 0 → false ⊦ ¬ by ∈ int2bytes(i1 / 256) ∧ by ≠ int2bytes((i1 + 255) % 256).last, int2bytes((i1 + 255) % 256).last ≠ by ∧ ¬ int2bytes((i1 + 255) % 256).last ∈ bytes

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ by ∈ by0 ' ↔ by = by0;% ⊦ by ∈ bytes + bytes0 ↔ ¬ ¬(by ∈ bytes ∨ by ∈ bytes0);% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ by ' + bytes = bytes0 → by ∈ bytes0; (forward)% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + by ' = bytes0 → by ∈ bytes0; (forward)%j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* j0);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;