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Proof step 8 from proof int2bytes-last-01 (previous step)

i < 128 ∨ 256 ≤ i, i < ~128 ∨ 128 ≤ i ⊦ (¬ i ≤ ~256 ∧ ¬ ~128 ≤ i ⊃ i2b(~1) ' + i2b(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 9:
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)

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ (bytes + by ').last = by;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;