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Proof step 23 from proof int2bytes-first-neg (previous step)

bytes = bytes0 + by0 ', int2bytes(i1 / 256) = by ' + bytes0, int2bytes((i1 + 255) % 256).last = by0, 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 ⊦ bytes = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
int2bytes(i1 / 256) = by ' + bytes0, 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 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + by ' ≠ [];