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Proof step 2 from proof bytes2int-int2bytes-len-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# bytes = # int2bytes(bytes2int(bytes)), ∀ bytes0. # bytes0 < # bytes → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ bytes = int2bytes(bytes2int(bytes))

Proof step 2 from proof bytes2int-int2bytes-len-eq (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information