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Proof step 5 from proof bytes2int-int2bytes-len-eq (previous step)

¬ 1 < # bytes, # bytes = # int2bytes(bytes2int(bytes)), ∀ bytes0. # bytes0 < # bytes → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ bytes = int2bytes(bytes2int(bytes))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
1 = # int2bytes(bytes2int(bytes)), # bytes = 1, ∀ bytes0. # bytes0 < 1 → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ bytes = int2bytes(bytes2int(bytes)), bytes = []

Rule argument

Additional information

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