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Proof step 17 from proof int2bytes-first-neg-01 (previous step)

i = i1 + 255, int2bytes(i1 / 256) + int2bytes(i % 256).last ' = i2b(0) ' + bytes, i < ~128, i < 0, ∀ bytes, i0. i→n(i0) < i→n(i) → int2bytes(i0) = i2b(0) ' + bytes ∧ i0 < 0 ∧ i2b(0) ∈ int2bytes(i0) → false ⊦ int2bytes(i % 256).last ≠ i2b(0) ∧ ¬ int2bytes(i % 256).last ∈ bytes, ¬ i2b(0) ∈ int2bytes(i1 / 256) ∧ i2b(0) ≠ int2bytes(i % 256).last

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
int2bytes(i1 / 256) + int2bytes((i1 + 255) % 256).last ' = i2b(0) ' + bytes, i1 < 0, i1 < ~383, ∀ bytes, i0. i→n(i0) < i→n(i1 + 255) → int2bytes(i0) = i2b(0) ' + bytes ∧ i0 < 0 ∧ i2b(0) ∈ int2bytes(i0) → false ⊦ int2bytes((i1 + 255) % 256).last ≠ i2b(0) ∧ ¬ int2bytes((i1 + 255) % 256).last ∈ bytes, ¬ i2b(0) ∈ int2bytes(i1 / 256) ∧ i2b(0) ≠ int2bytes((i1 + 255) % 256).last

Rule argument

Additional information

Used simplifier rules: ⊦ 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;