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Proof step 14 from proof int2bytes-len-one (previous step)

(i < ~256 ∧ i % 256 = 0 ⊃ int2bytes(i / 256) + i2b(i % 256) ' ;(i < ~256 ∧ i % 256 ≠ 0 ⊃ int2bytes(i / 256 - 1) + i2b(i % 256) ';int2bytes(i / 256) + i2b(i % 256) ') ) = by ', i ≤ ~256, i < 0 ⊦ i = b2i(by)

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) ≠ [];%int2bytes-byte: ~128 ≤ i, i < 128 ⊦ int2bytes((* i)) = i2b(i) ';%int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ by ' = by0 ' + bytes ↔ by = by0 ∧ bytes = [];% ⊦ by ' ≠ [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by ' = bytes + by0 ' ↔ by = by0 ∧ bytes = [];% ⊦ i ≠ j → (i ≤ j ↔ i < j);