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Proof step 17 from proof int2bytes-index-rightshift-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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