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Proof step 9 from proof int2bytes-last-01 (previous step)

i ≤ ~256 ∨ ~128 ≤ i, i < 128 ∨ 256 ≤ i, i < ~128 ∨ 128 ≤ i ⊦ (¬ ~256 ≤ i ∧ i % 256 = 0 ⊃ int2bytes(i / 256) + int2bytes(i % 256).last ' ;(¬ ~256 ≤ i ∧ i % 256 ≠ 0 ⊃ int2bytes(i / 256 - 1) + int2bytes(i % 256).last ';int2bytes(i / 256) + int2bytes(i % 256).last ') ) .last = i2b(i)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 10:
¬ ~256 ≤ i ∧ i % 256 = 0, i ≤ ~256 ∨ ~128 ≤ i, i < 128 ∨ 256 ≤ i, i < ~128 ∨ 128 ≤ i ⊦ (int2bytes(i / 256) + int2bytes(i % 256).last ').last = i2b(i)
Proof step 13:
~256 ≤ i ∨ i % 256 ≠ 0, i ≤ ~256 ∨ ~128 ≤ i, i < 128 ∨ 256 ≤ i, i < ~128 ∨ 128 ≤ i ⊦ (¬ ~256 ≤ i ∧ i % 256 ≠ 0 ⊃ int2bytes(i / 256 - 1) + int2bytes(i % 256).last ';int2bytes(i / 256) + int2bytes(i % 256).last ').last = i2b(i)

Rule argument

pos: right 1

Proof step 9 from proof int2bytes-last-01 (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information