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Proof step 9 from proof bytes2int-diff (previous step)

bytes2int(bytes0) = bytes2int(bytes), # bytes0 = 1, bytes = bytes.first + [], ∀ bytes0, bytes1. # bytes1 < 1 → # bytes0 = # bytes1 ∧ bytes2int(bytes0) = bytes2int(bytes1) → bytes0 = bytes1 ⊦ bytes0 = bytes, bytes = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
bytes2int(bytes0) = bytes2int(bytes), # bytes0 = 1, bytes = bytes.first ', ∀ bytes0, bytes1. # bytes1 < 1 → # bytes0 = # bytes1 ∧ bytes2int(bytes0) = bytes2int(bytes1) → bytes0 = bytes1 ⊦ bytes0 = bytes, bytes = []

Rule argument

Additional information

Used simplifier rules: ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + [] = bytes;% ⊦ by + bytes = by ' + bytes;