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Proof step 1 from proof int2bytes-len-ls-03 (conclusion)

⊦ 0 ≤ i ∧ int2bytes(i).first ≠ i2b(0) → (# int2bytes(i) < n ↔ i < 2 ^ ((n→i(n) - 1) * 8))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ i ⊦ int2bytes(i).first = i2b(0), # int2bytes(i) < n ↔ ¬ 2 ^ ((n→i(n) - 1) * 8) ≤ i

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;