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Proof step 15 from proof int2bytes-rec-case (previous step)

i < 0, i ≤ ~256 ∨ ~128 ≤ i, i < 128 ∨ 256 ≤ i, i < ~128 ∨ 128 ≤ i ⊦ int2bytes(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 ') )

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
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 ') )

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;% ⊦ j ≤ i ∧ i < 0 → j < 0; (forward)% ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)%j0 ≠ 0, j < j0 ⊦ i ≤ (* j) → i < (* j0);% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ i < 0 ∧ 0 < j → ¬ j ≤ i;%j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* j0);