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Proof step 12 from proof int2bytes-len-one (previous step)

(~256 < i ⊃ i2b(~1) ' + i2b(i) ' ;(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 < ~128, i < 0 ⊦ i = b2i(by)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 13:
¬ ~256 < i, (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 < ~128, i < 0 ⊦ i = b2i(by)

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ by ' = by0 ' + bytes ↔ by = by0 ∧ bytes = [];% ⊦ by ' ≠ [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;