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Proof step 2 from proof int2bytes-len-one-01 (
previous step
)
⊦ # int2bytes(i) = 1 ↔ i = b2i(i2b(i))
User applied rule case distinction
And obtained the following new goals
Proof step 3
:
# int2bytes(i) = 1 ⊦ i = b2i(i2b(i))
Proof step 16
:
i = b2i(i2b(i)) ⊦ # int2bytes(i) = 1
Rule argument
pos: right 1
Additional information