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Proof step 4 from proof cast-one (previous step)

⊦ bits2int(lastn(i→n(1), fillsgn(i→n(1), int2bits(j)))) = 0, bits2int(lastn(i→n(1), fillsgn(i→n(1), int2bits(j)))) = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ bits2int(fillsgn(1, int2bits(j)).last ') = 0, bits2int(fillsgn(1, int2bits(j)).last ') = ~1

Rule argument

Additional information

Used simplifier rules: ⊦ lastn(1, bits) = bits.last ';