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Proof step 6 from proof bint-is-cast (
previous step
)
¬ i ∈bint ⊦ bi2i(i2bi(bits2int(lastn(32, int2bits(i))))) = cast(32, i)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 7
:
⊦ bi2i(i2bi(bits2int(lastn(32, int2bits(i))))) = cast(32, i), i ∈bint
Rule argument
Additional information