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Proof step 13 from proof cast-law-and-one (previous step)

0 < i ⊦ bits2int(lastn(i→n(i), fillsgn(i→n(i), bitsand(int2bits(j), int2bits(j0))))) = cast(i, bits2int(bitsand(int2bits(bits2int(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: bits ≠ [] ⊦ int2bits(bits2int(bits)) = sgn(bits);% ⊦ lastn(n, bits) = [] ↔ n = 0;% ⊦ i→n(i) = 0 ↔ i = 0;