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Proof step 1 from proof int2int-lastn-pospos (conclusion)

⊦ 0 < i ∧ ¬ lastn(m, int2bits(i)).first ∧ m ≤ # int2bits(i) ∧ m ≠ 0 → bits2int(lastn(m, int2bits(i))) = i % 2 ^ n→i(m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information