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Proof step 27 from proof int2bytes-index-rightshift (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 28:
0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n

Rule argument

seq: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n subst: [n, i] → [# int2bytes(i), j] rotate: F spec: intnat inst: name: num-n2i-ls-02

Additional information

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