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Proof step 19 from proof int2bytes-one (previous step)

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

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ #(bytes + bytes0) = # bytes + # bytes0;% ⊦ # by ' = 0 +1;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ 1 = m + 1 ↔ m = 0;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;