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Proof step 20 from proof int2bytes-index-rightshift-02 (previous step)

¬ i→n(i1) ≤ # int2bytes(i), 0 < i1, i1 ≤ 4, i1 ≤ n→i(# int2bytes(i)), 0 ≤ i1, # int2bytes(i) - i→n(i1) < # int2bytes(i) ⊦ 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: ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ m +1 = m + 1;% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward)% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;