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Proof step 8 from proof int2bytes-len-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) + i2b(i % 256) ' ;(i < ~256 ∧ i % 256 ≠ 0 ⊃ int2bytes(i / 256 - 1) + i2b(i % 256) ';int2bytes(i / 256) + i2b(i % 256) ') ) ) ) ) = by ' ⊦ i = b2i(by), ¬ i < ~128 ∧ ¬ 128 ≤ i

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ by ' = by0 ' ↔ by = by0;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);