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Proof step 10 from proof int2bytes-index-rightshift (previous step)

0 ≤ j, j < n→i(# int2bytes(i)), # int2bytes(i) ≤ # int2bytes(i) - i→n(j) + 1, # int2bytes(i) ≤ i→n(j) ⊦ 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 11:
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: ⊦ m ≤ n → (n < m ↔ false); Used terms: # int2bytes(i), j