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Proof step 19 from proof int2bytes-first-neg (previous step)

by ' + bytes ≠ [], int2bytes(i1 / 256) = (by ' + bytes).butlast ∧ int2bytes((i1 + 255) % 256).last = (by ' + bytes).last, 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
int2bytes(i1 / 256) = (by ' + bytes).butlast, int2bytes((i1 + 255) % 256).last = (by ' + bytes).last, 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 ≠ (by ' + bytes).last, (by ' + bytes).last ≠ by ∧ ¬ (by ' + bytes).last ∈ bytes

Rule argument

Additional information

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