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Proof step 6 from proof bytes2int-diff (previous step)

¬ 1 < # bytes, bytes2int(bytes0) = bytes2int(bytes), # bytes0 = # bytes, ∀ bytes0, bytes1. # bytes1 < # bytes → # bytes0 = # bytes1 ∧ bytes2int(bytes0) = bytes2int(bytes1) → bytes0 = bytes1 ⊦ bytes0 = bytes

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
bytes2int(bytes0) = bytes2int(bytes), # bytes0 = 1, # bytes = 1, ∀ bytes0, bytes1. # bytes1 < 1 → # bytes0 = # bytes1 ∧ bytes2int(bytes0) = bytes2int(bytes1) → bytes0 = bytes1 ⊦ bytes0 = bytes, bytes = []

Rule argument

Additional information

Used simplifier rules: ⊦ # [] = 0;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);