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Proof step 2 from proof int2bytes-index-rightshift-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
j ≤ i, 0 ≤ j ⊦ i→n(i - j) = i→n(i) - i→n(j)
Proof step 4:
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))

Rule argument

seq: j ≤ i, 0 ≤ j ⊦ i→n(i - j) = i→n(i) - i→n(j) subst: [j, i] → [i1, n→i(# int2bytes(i))] rotate: F spec: intnat inst: name: sub

Additional information

Used simplifier rules: ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ i < j → ¬ j < i; Used terms: i1, n→i(# int2bytes(i))