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Proof step 42 from proof rightshift-8 (previous step)

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

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: int2int: ⊦ bytes2int(int2bytes(i)) = i;%int2bytes-empty: ⊦ int2bytes(i) ≠ [];%bytes2int-one: ⊦ bytes2int(by ') = b2i(by);%~128 ≤ i, i ≤ 127 ⊦ b2i(i2b((* i))) = i;% ⊦ (bytes + by ').butlast = bytes;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ i + (j - i) = j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;% ⊦ # by ' = 0 +1;% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;