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Proof step 14 from proof int2bytes-index-rightshift (previous step)

i→n(n→i(# int2bytes(i)) - j + 1) < # int2bytes(i), 0 ≤ j, j < n→i(# int2bytes(i)) ⊦ int2bytes(i)[i→n(j)] = i2b(bytes2int(firstn(# int2bytes(i) - i→n(n→i(# int2bytes(i)) - j + 1), int2bytes(i))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
j < n→i(# int2bytes(i)), 0 ≤ j, i→n(n→i(# int2bytes(i)) - j + 1) < # int2bytes(i) ⊦ int2bytes(i)[i→n(j)] = i2b(bytes2int(firstn(# int2bytes(i) - i→n(n→i(# int2bytes(i)) - j + 1), int2bytes(i))))

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m < n → n ≠ 0; (forward)