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Proof step 9 from proof int2bytes-len-one-01 (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) ') ) ) ) ) = 1 ⊦ i = b2i(i2b(i)), ¬ i < ~128 ∧ ¬ 127 < 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 ' = 0 +1;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);%i + 1 = i0 ⊦ j < (* i0) → ¬ (* i) < j;