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Proof step 40 from proof int2bytes-leading-zeros (previous step)

¬(127 < i / 256 ∧ i / 256 ≤ 255), int2bytes(i / 256) = i2b(0) ', int2bytes(i % 256) = bytes0 + i2b(0) ', i / 256 ≤ 255, i2b(0) ∈ int2bytes(i % 256), 0 < i, 0 ≤ i, 255 < 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 41:
int2bytes(i / 256) = i2b(0) ', int2bytes(i % 256) = bytes0 + i2b(0) ', i2b(0) ∈ int2bytes(i % 256), 0 < i, 0 ≤ i, 255 < i, i / 256 ≤ 127, ∀ 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: j0 ≠ 0, j ≤ j0 ⊦ i ≤ (* j) → i ≤ (* j0);% ⊦ ¬ i < j ↔ j ≤ i;