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Proof step 4 from proof int2bytes-last-01 (previous step)

⊦ (¬ i < ~128 ∧ ¬ 128 ≤ i ⊃ i2b(i) ' ;(¬ i < 128 ∧ ¬ 256 ≤ i ⊃ i2b(0) ' + i2b(i - 256) ' ;(¬ 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

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ by '.last = by;