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Proof step 10 from proof int2bytes-not-neg (previous step)

¬ i < 256, b2i((int2bytes(i / 256) + i2b(i % 256) ').first) < 0, 0 < i, 128 ≤ i, 0 ≤ i, ∀ 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 11:
b2i((int2bytes(i / 256) + i2b(i % 256) ').first) < 0, 256 ≤ i, 0 ≤ i, 128 ≤ i, 0 < i, ∀ i0. i→n(i0) < i→n(i) → b2i(int2bytes(i0).first) < 0 ∧ 0 ≤ i0 → false ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ ¬ i < j ↔ j ≤ i;