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Proof step 23 from proof int2bytes-index-rightshift (previous step)

j + 1 ≤ n→i(# int2bytes(i)) ∧ 0 ≤ j + 1, j < n→i(# int2bytes(i)), 0 ≤ j, i→n(n→i(# int2bytes(i))) - i→n(j + 1) < # int2bytes(i) ⊦ int2bytes(i)[i→n(j)] = int2bytes(i)[# int2bytes(i) - (i→n(n→i(# int2bytes(i))) - i→n(j + 1)) - 1]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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