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Proof step 4 from proof int2bytes-butlast (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i < ~128 ∨ 128 ≤ i ⊦ (¬ 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) + 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)

Rule argument

Additional information

Used simplifier rules: int2bytes-last-01: ⊦ int2bytes(i).last = i2b(i);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;