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Proof step 27 from proof rightshift-8 (previous step)

¬ i ≤ ~256, i % 256 = 0, i < ~128, i < 0 ⊦ i / 256 = bytes2int((i2b(~1) ' + i2b(i) ').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 28:
i % 256 = 0, i < ~128, i < 0, ~256 < i ⊦ i / 256 = ~1

Rule argument

Additional information

Used simplifier rules: bytes2int-one: ⊦ bytes2int(by ') = b2i(by);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;%~128 ≤ i, i ≤ 127 ⊦ b2i(i2b((* i))) = i;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ (bytes + by ').butlast = bytes;