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Proof step 22 from proof bytes2int-fillsgn-zero (previous step)

# bytes = m + 1, n ≤ m, ∀ by. by ∈ bytes → b2i(by) = 0 ⊦ bytes = [], m = 0, (mklist(i2b(0), m) + i2b(0) ')[n] = bytes[n]

User applied rule all left

And obtained the following new goals

Proof step 23:
# bytes = m + 1, n ≤ m, bytes[n] ∈ bytes → b2i(bytes[n]) = 0 ⊦ bytes = [], m = 0, (mklist(i2b(0), m) + i2b(0) ')[n] = bytes[n]

Rule argument

pos: left 3 terms: bytes[n] abort: F discard: T computed: F

Additional information

Used terms: bytes[n]