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Proof step 24 from proof int2bytes-leading-zeros (previous step)

¬(~256 < i1 ∧ i1 < ~128), b2i(i2b(i1)) = 0, ~256 ≤ i1, ~129 < i1, i1 < 0, ∀ bytes, i0. i→n(i0) < i→n(i1 + 256) → int2bytes(i0) = i2b(0) ' + i2b(0) ' + bytes ∧ 0 ≤ i0 → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
b2i(i2b(i1)) = 0, i1 < 0, ~128 ≤ i1, ∀ bytes, i0. i→n(i0) < i→n(i1 + 256) → int2bytes(i0) = i2b(0) ' + i2b(0) ' + bytes ∧ 0 ≤ i0 → false ⊦

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 < i ⊦ (* i) ≤ j → (* i0) < j;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ ¬ i < j ↔ j ≤ i;%i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;