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Proof step 14 from proof intshiftrsign32 (previous step)

# fillsgn(32, int2bits(i)) = 32, band(j, 31) = 0, # int2bits(i) ≤ 32 ⊦ i = bits2int(fillsgn(32, int2bits(i)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: fill-length: ⊦ # bits ≤ n → # fillsgn(n, bits) = n; from specification bitopsaux