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Proof step 23 from proof cast-law-and-one (previous step)

lastn(i→n(i), fillsgn(i→n(i), int2bits(j))) ≠ [] ∧ int2bits(j0) ≠ [], bitsand(sgn(lastn(i→n(i), fillsgn(i→n(i), int2bits(j)))), int2bits(j0)) ≠ [], 0 < i ⊦ bits2int(lastn(i→n(i), fillsgn(i→n(i), bitsand(int2bits(j), int2bits(j0))))) = bits2int(lastn(i→n(i), fillsgn(i→n(i), sgn(bitsand(lastn(i→n(i), fillsgn(i→n(i), int2bits(j))), int2bits(j0))))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ lastn(n, bits) = [] ↔ n = 0;% ⊦ int2bits(j) ≠ [];