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Proof step 8 from proof bytes2int-fillsgn-zero (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 9:
# bytes = 1, bytes.first ∈ bytes → b2i(bytes.first) = 0 ⊦ bytes = [], fillsgn(1, i2b(0) ') = bytes

Rule argument

pos: left 2 terms: bytes.first abort: F discard: T computed: F

Additional information

Used terms: bytes.first