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Proof step 11 from proof int2bytes-one (previous step)

# (~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 ') ) ) ) ) = 1, i < ~128, i < 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# (~256 < i ⊃ i2b(~1) ' + i2b(i) ' ;(i < ~256 ∧ i % 256 = 0 ⊃ int2bytes(i / 256) + i2b(i % 256) ' ;(i < ~256 ∧ i % 256 ≠ 0 ⊃ int2bytes(i / 256 - 1) + i2b(i % 256) ';int2bytes(i / 256) + i2b(i % 256) ') ) ) = 1, i < ~128, i < 0 ⊦

Rule argument

Additional information

Used simplifier rules: int2bytes-last-01: ⊦ int2bytes(i).last = i2b(i);%j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0);% ⊦ i < 0 ∧ 0 < j → ¬ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);