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Proof step 18 from proof int2bytes-index-rightshift-02 (previous step)

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))

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ n ≤ m → n→i(m - n) = n→i(m) - n→i(n) subst: [n, m] → [i→n(i1), # int2bytes(i)] rotate: F spec: intnat inst: name: n2i-sub

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ 0 < i → n→i(i→n(i)) = i;% ⊦ i + (j - i0) = i + j - i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i - (j - j0) = i + j0 - j;% ⊦ i + j - i + 1 = j - 1; Used terms: i→n(i1), # int2bytes(i)