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Proof step 18 from proof int2bytes-leading-zeros (previous step)

i = i1 + 256, i2b(i1) = i2b(0), 0 < i, i ≤ 255, 0 ≤ i, 127 < i, ∀ bytes, i0. i→n(i0) < i→n(i) → int2bytes(i0) = i2b(0) ' + i2b(0) ' + bytes ∧ 0 ≤ i0 → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)%i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ i ≤ ~1 ↔ i < 0;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;