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Proof step 20 from proof intshiftrsign32 (previous step)

# int2bits(i) ≤ 32 ∧ int2bits(i) ≠ [], 32 = n0 + i→n(band(j, 31)), # int2bits(i) ≤ 32, band(j, 31) ≤ 32 ⊦ bits2int(sublist(0, # int2bits(i), mklist(¬ 0 ≤ i, i→n(band(j, 31))) + int2bits(i))) = bits2int((¬ 0 ≤ i)' + sublist(0, n0, mklist(int2bits(i).first, 32 - # int2bits(i)) + int2bits(i))), band(j, 31) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
32 = n0 + i→n(band(j, 31)), # int2bits(i) ≤ 32, band(j, 31) ≤ 32 ⊦ band(j, 31) = 0, bits2int(sublist(0, # int2bits(i), mklist(¬ 0 ≤ i, i→n(band(j, 31))) + int2bits(i))) = bits2int((¬ 0 ≤ i)' + sublist(0, n0, mklist(¬ 0 ≤ i, 32 - # int2bits(i)) + int2bits(i)))

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ []; from specification int2bits%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%int2bits-first-02: ⊦ int2bits(j).first ↔ j < 0; from specification int2bits