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Proof step 12 from proof int2bytes-len-one-01 (previous step)

¬ 128 ≤ i ∨ ¬ i < 256, # (~256 < i ∧ i < ~128 ⊃ 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) ') ) ) = 1, i < ~128 ∨ 128 ≤ i ⊦ i = b2i(i2b(i)), ¬ i < ~128 ∧ ¬ 127 < i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# (~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) ') ) ) = 1, i < ~128, i < 0 ⊦ i = b2i(i2b(i))

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)%j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0);%j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* j0);% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ 1 = m + 1 ↔ m = 0;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;% ⊦ i < 0 ∧ 0 ≤ j → ¬ j < i;% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;