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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ ¬ i < j ↔ j ≤ i;