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Proof step 20 from proof int2bytes-rec-case (previous step)

i = ~256 ∨ i % 256 ≠ 0, i < 0, i ≤ ~256 ⊦ int2bytes(i) = (i ≠ ~256 ∧ i % 256 ≠ 0 ⊃ int2bytes(i / 256 - 1) + int2bytes(i % 256).last ';int2bytes(i / 256) + int2bytes(i % 256).last ')

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
i < 0, i < ~256 ⊦ i = ~256, i % 256 = 0, int2bytes(i) = int2bytes(i / 256 - 1) + int2bytes(i % 256).last '

Rule argument

Additional information

Used simplifier rules: int2bytes-minus-256: ⊦ int2bytes(~256) = i2b(~1) ' + i2b(0) ';%int2bytes-byte: ~128 ≤ i, i < 128 ⊦ int2bytes((* i)) = i2b(i) ';% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes + bytes1 ↔ bytes0 = bytes1;% ⊦ by '.last = by;