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Proof step 9 from proof int2bytes-index-rightshift-02 (previous step)

0 ≤ 8 * (i1 - 1) ∧ 8 * (i1 - 1) ≤ 31, 0 ≤ i1, i1 ≤ n→i(# int2bytes(i)), i1 ≤ 4, 0 < i1 ⊦ int2bytes(i)[# int2bytes(i) - i→n(i1)] = i2b(i >> 8 * (i1 - 1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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