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Proof step 34 from proof int2bytes-leading-zeros (previous step)

int2bytes(i % 256) = bytes0 + by ' ∧ int2bytes(i % 256) ≠ [], by = i2b(0), int2bytes(i / 256) = i2b(0) ', 255 < i, 0 ≤ i, 0 < i, ∀ bytes, i0. i→n(i0) < i→n(i) → int2bytes(i0) = i2b(0) ' + i2b(0) ' + bytes ∧ 0 ≤ i0 → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
int2bytes(i % 256) = bytes0 + i2b(0) ', int2bytes(i / 256) = i2b(0) ', 255 < i, 0 ≤ i, 0 < i, i2b(0) ∈ int2bytes(i % 256), ∀ bytes, i0. i→n(i0) < i→n(i) → int2bytes(i0) = i2b(0) ' + i2b(0) ' + bytes ∧ 0 ≤ i0 → false ⊦

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + by ' = bytes0 → by ∈ bytes0; (forward)