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Proof step 4 from proof int2int-lastn-2pot (previous step)

2 ^ n→i(m) ≠ 0, ∃ j0. i = j0 * 2 ^ n→i(m), m ≤ # int2bits(i) ⊦ m = 0, bits2int(lastn(m, int2bits(i))) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m ≤ # int2bits(j * 2 ^ n→i(m)) ⊦ m = 0, bits2int(lastn(m, int2bits(j * 2 ^ n→i(m)))) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ 2 ^ n→i(m) ≠ 0;