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Proof step 49 from proof int2bytes-leading-zeros (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 50:
int2bytes(i / 256) = i2b(0) ' + i2b(0) ' + bytes.butlast, int2bytes(i % 256).last = bytes.last, 0 < i, 0 ≤ i, 255 < i, ∀ bytes, i0. i→n(i0) < i→n(i) → int2bytes(i0) = i2b(0) ' + i2b(0) ' + bytes ∧ 0 ≤ i0 → false ⊦ bytes = []

Rule argument

Additional information

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