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Proof step 13 from proof int2bytes-len-one (previous step)

¬ ~256 < 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) ') ) = by ', i < ~128, i < 0 ⊦ i = b2i(by)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
(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) ') ) = by ', i ≤ ~256, i < 0 ⊦ i = b2i(by)

Rule argument

Additional information

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