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

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

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information