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Proof step 16 from proof int2bytes-rec-case (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 17:
i ≠ ~256 ∧ i % 256 = 0, i < 0, i ≤ ~256 ⊦ int2bytes(i) = int2bytes(i / 256) + int2bytes(i % 256).last '
Proof step 20:
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 ')

Rule argument

pos: right 1

Proof step 16 from proof int2bytes-rec-case (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information