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Proof step 5 from proof bits2int-neg (previous step)

0 ≤ bytes2int(bits2bytes(bits)) ↔ ¬ bits.first, bytes2int(bits2bytes(bits)) < 0 ⊦ bits = [], bits.first

User applied rule case distinction

And obtained the following new goals

Proof step 6:
0 ≤ bytes2int(bits2bytes(bits)), ¬ bits.first, bytes2int(bits2bytes(bits)) < 0 ⊦ bits = [], bits.first

Rule argument

pos: left 1

Additional information