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Proof step 4 from proof int2bytes-index-rightshift-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ n→i(n);% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i→n(n→i(n)) = n;