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Proof step 24 from proof int2bytes-index-rightshift (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ n ≤ m → m - (m - n) = n
Proof step 26:
¬ i→n(j) + 1 ≤ # int2bytes(i), j < n→i(# int2bytes(i)), 0 ≤ j, # int2bytes(i) - i→n(j) + 1 < # int2bytes(i) ⊦ int2bytes(i)[i→n(j)] = int2bytes(i)[# int2bytes(i) - (# int2bytes(i) - i→n(j) + 1) - 1]

Rule argument

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

Additional information

Used simplifier rules: ⊦ n ≤ m → m + n0 - n = (m - n) + n0;% ⊦ 0 + m = m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: i→n(j) + 1, # int2bytes(i)