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

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

Rule argument

Additional information

Used simplifier rules: ⊦ i - j = i0 ↔ i = i0 + j;