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Proof step 21 from proof int2bytes-first-neg (previous step)

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 ⊦ bytes = [], ¬ by ∈ int2bytes(i1 / 256) ∧ by ≠ (by ' + bytes).last, (by ' + bytes).last ≠ by ∧ ¬ (by ' + bytes).last ∈ bytes

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
int2bytes(i1 / 256) = by ' + bytes.butlast, int2bytes((i1 + 255) % 256).last = 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 ⊦ bytes = []

Rule argument

Additional information

Used simplifier rules: ⊦ by ∈ by0 ' ↔ by = by0;% ⊦ by ∈ bytes + bytes0 ↔ ¬ ¬(by ∈ bytes ∨ by ∈ bytes0);% ⊦ bytes.last ∈ bytes ↔ bytes ≠ [];% ⊦ by ' + bytes = bytes0 → by ∈ bytes0; (forward)% ⊦ bytes ≠ [] → (by ' + bytes).last = bytes.last;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes ≠ [] → (bytes0 + bytes).butlast = bytes0 + bytes.butlast;