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Proof step 4 from proof int2int-lastn-negpos (previous step)

∃ bits0. int2bits(i) = bits0 + lastn(m, int2bits(i)) ∧ # bits0 = # int2bits(i) - m, 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), i < 0, m ≤ # int2bits(i) ⊦ bits2int(lastn(m, int2bits(i))) = i % 2 ^ n→i(m) + 2 ^ n→i(m), i % 2 ^ n→i(m) = 0, m = 0, lastn(m, int2bits(i)).first

Proof step 4 from proof int2int-lastn-negpos (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information