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Proof step 1 from proof int2bytes-index-rightshift-02 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 < i1, i1 ≤ 4, i1 ≤ n→i(# int2bytes(i)) ⊦ int2bytes(i)[i→n(n→i(# int2bytes(i)) - i1)] = i2b(i i>> 8 * (i1 - 1))

Rule argument

Additional information

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