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Proof step 1 from proof bytes2int-size-val-03 (conclusion)

⊦ ∃ by, bytes. 0 < b2i(by) ∧ bytes2int(by ' + bytes) = 2 ^ (n→i(# bytes) * 8)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ ∃ by, bytes. ¬ b2i(by) ≤ 0 ∧ bytes2int(by ' + bytes) = 2 ^ (n→i(# bytes) * 8)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;