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Proof step 6 from proof int2bytes-not-neg (previous step)

0 ≤ i, b2i ((~128 ≤ i ∧ i < 128 ⊃ i2b(i) ' ;(128 ≤ i ∧ i < 256 ⊃ i2b(0) ' + i2b(i - 256) ' ;(~256 < i ∧ i < ~128 ⊃ i2b(~1) ' + i2b(i) ' ;(i < ~256 ∧ i % 256 = 0 ⊃ int2bytes(i / 256) + int2bytes(i % 256).last ' ;(i < ~256 ∧ i % 256 ≠ 0 ⊃ int2bytes(i / 256 - 1) + int2bytes(i % 256).last ';int2bytes(i / 256) + int2bytes(i % 256).last ') ) ) ) ) .first) < 0, ∀ i0. i→n(i0) < i→n(i) → b2i(int2bytes(i0).first) < 0 ∧ 0 ≤ i0 → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
b2i((i < 128 ⊃ i2b(i) ';(128 ≤ i ∧ i < 256 ⊃ i2b(0) ' + i2b(i - 256) ';int2bytes(i / 256) + i2b(i % 256) ')).first) < 0, 0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → b2i(int2bytes(i0).first) < 0 ∧ 0 ≤ i0 → false ⊦

Rule argument

Additional information

Used simplifier rules: int2bytes-last-01: ⊦ int2bytes(i).last = i2b(i);% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i < 0 ∧ 0 ≤ j → ¬ j < i;%i0 ≠ 0, i0 < i ⊦ (* i) ≤ j → (* i0) < j;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;