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Proof step 38 from proof rightshift-8 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
i < ~128, i < 0, ~256 < i ⊦ i % 256 = 0, i / 256 = 0

Rule argument

Additional information

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