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Proof step 1 from proof bytes2int-size-val-05 (conclusion)

⊦ ∃ by, bytes. 0 ≤ b2i(by) ∧ ~(2 ^ ((n→i(# bytes) + 1) * 8)) = bytes2int(i2b(~1) ' + by ' + bytes)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ ∃ by, bytes. ¬ b2i(by) < 0 ∧ 2 ^ ((n→i(# bytes) + 1) * 8) + bytes2int(i2b(~1) ' + by ' + bytes) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ~ i = j ↔ i + j = 0;% ⊦ ¬ i ≤ j ↔ j < i;