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

∃ bits0. int2bits(i) = bits0 + lastn(m, int2bits(i)) ∧ # bits0 = # int2bits(i) - m, m ≤ # int2bits(i), 0 < i ⊦ m = 0, bits2int(lastn(m, int2bits(i))) = i % 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)), 0 < i, m ≤ # int2bits(i) ⊦ bits2int(lastn(m, int2bits(i))) = i % 2 ^ n→i(m), m = 0, lastn(m, int2bits(i)).first

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information