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Proof step 26 from proof int2bytes-butlast (previous step)

i ≤ ~256 ∨ ~128 ≤ i, i < 128 ∨ 256 ≤ i, i < ~128 ∨ 128 ≤ i, i < ~128 ∨ 128 ≤ i ⊦ (¬ ~256 ≤ i ∧ i % 256 = 0 ⊃ int2bytes(i / 256) + i2b(i % 256) ' ;(¬ ~256 ≤ i ∧ i % 256 ≠ 0 ⊃ int2bytes(i / 256 - 1) + i2b(i % 256) ';int2bytes(i / 256) + i2b(i % 256) ') ) .butlast = int2bytes(i >> 8)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 27:
¬ ~256 ≤ i ∧ i % 256 = 0, i ≤ ~256 ∨ ~128 ≤ i, i < 128 ∨ 256 ≤ i, i < ~128 ∨ 128 ≤ i, i < ~128 ∨ 128 ≤ i ⊦ (int2bytes(i / 256) + i2b(i % 256) ').butlast = int2bytes(i >> 8)
Proof step 33:
~256 ≤ i ∨ i % 256 ≠ 0, i ≤ ~256 ∨ ~128 ≤ i, i < 128 ∨ 256 ≤ i, i < ~128 ∨ 128 ≤ i, i < ~128 ∨ 128 ≤ i ⊦ (¬ ~256 ≤ i ∧ i % 256 ≠ 0 ⊃ int2bytes(i / 256 - 1) + i2b(i % 256) ';int2bytes(i / 256) + i2b(i % 256) ').butlast = int2bytes(i >> 8)

Rule argument

pos: right 1

Proof step 26 from proof int2bytes-butlast (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information