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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ ~ i = j ↔ i + j = 0;