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Proof step 23 from proof int2bytes-rec-case (previous step)

i ≤ ~256 ∨ ~128 ≤ i, i < 128 ∨ 256 ≤ i, i < ~128 ∨ 128 ≤ i ⊦ i < 0, 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 24:
256 ≤ i, 0 < i, 0 ≤ i ⊦ int2bytes(i) = int2bytes(i / 256) + int2bytes(i % 256).last '

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ i < 0 ∧ 0 ≤ j → ¬ j < i;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ ¬ i < j ↔ j ≤ i;