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Proof step 13 from proof rightshift-8 (previous step)

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

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

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