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Proof step 5 from proof int2bytes-index-rightshift (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
j ≤ i, 0 ≤ j ⊦ i→n(i - j) = i→n(i) - i→n(j)
Proof step 7:
j + 1 ≤ n→i(# int2bytes(i)) ∧ 0 ≤ j + 1, 0 ≤ j, j < n→i(# int2bytes(i)), # int2bytes(i) ≤ i→n(n→i(# int2bytes(i))) - i→n(j + 1) ⊦ int2bytes(i)[i→n(j)] = i2b(i >> 8 * (n→i(# int2bytes(i)) - j + 1))

Rule argument

seq: j ≤ i, 0 ≤ j ⊦ i→n(i - j) = i→n(i) - i→n(j) subst: [j, i] → [j + 1, n→i(# int2bytes(i))] rotate: F spec: intnat inst: name: sub

Additional information

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