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Proof step 1 from proof int2int-lastn-2pot (conclusion)
⊦ i % 2 ^ n→i(m) = 0 ∧ m ≤ # int2bits(i) ∧ m ≠ 0 → bits2int(lastn(m, int2bits(i))) = 0
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 2
:
i % 2 ^ n→i(m) = 0, m ≤ # int2bits(i) ⊦ m = 0, bits2int(lastn(m, int2bits(i))) = 0
Rule argument
Additional information