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Proof step 9 from proof bytes2int-fillsgn-zero (previous step)

# bytes = 1, bytes.first ∈ bytes → b2i(bytes.first) = 0 ⊦ bytes = [], fillsgn(1, i2b(0) ') = bytes

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# bytes = 1, b2i(bytes.first) = 0 ⊦ bytes = [], fillsgn(1, i2b(0) ') = bytes

Rule argument

Additional information

Used simplifier rules: ⊦ bytes.first ∈ bytes ↔ bytes ≠ [];