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Proof step 30 from proof rightshift-8 (previous step)

i ≤ ~256, i % 256 = 0, i < ~128, i < 0 ⊦ i / 256 = bytes2int((i ≠ ~256 ⊃ int2bytes(i / 256) + i2b(0) ';i2b(~1) ' + i2b(0) ').butlast), # (¬ i ≤ ~256 ⊃ i2b(~1) ' + i2b(i) ';(i ≠ ~256 ⊃ int2bytes(i / 256) + i2b(0) ';i2b(~1) ' + i2b(0) ')) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
i % 256 = 0, i < 0, i ≤ ~256 ⊦ # (i ≠ ~256 ⊃ int2bytes(i / 256) + i2b(0) ';i2b(~1) ' + i2b(0) ') = 1, i / 256 = bytes2int((i ≠ ~256 ⊃ int2bytes(i / 256) + i2b(0) ';i2b(~1) ' + i2b(0) ').butlast)

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);%j0 ≠ 0, j ≤ j0 ⊦ i ≤ (* j) → i ≤ (* j0);