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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
m ≤ # int2bits(i), i < 0 ⊦ m = 0, i % 2 ^ n→i(m) = 0, bits2int(lastn(m, int2bits(i))) = i % 2 ^ n→i(m) + 2 ^ n→i(m), lastn(m, int2bits(i)).first

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information